If $$\sum_{r=1}^{n}(r)(r+1)(2r+3)=an^4+bn^3+cn^2+dn+e$$, then

$$a, b-\dfrac23, c-1$$ are in A.P.

MCQ Questions for Class 11 Engineering Maths Sequences And Series Quiz 11

Let

$\displaystyle a_{1},a_{2},a_{3},....,a_{11}$ be real numbers satisfying

$\displaystyle a_{1}=15,27-2a_{2}> 0\:$ and

$\: \: a_{k}=2a_{k-1}-a_{k-2}$ for

$k = 3, 4, ......., 11$ .

$\displaystyle \frac{a_{1}^{2}+a_{2}^{2}+...+a_{11}^{2}}{11}=90$ , then the value of

$\displaystyle \frac{a_{1}+a_{2}+...+a_{11}}{11}$ is equal to

0%
1
0%
5
0%
9
0%
0

Explanation

${ a }_{ 1 }=15\\ { a }_{ 3 }=2{ a }_{ 2 }-{ a }_{ 1 }\\ { a }_{ 4 }=2{ a }_{ 3 }-{ a }_{ 2 }=4{ a }_{ 2 }-2{ a }_{ 1 }-{ a }_{ 2 }=3{ a }_{ 2 }-2{ a }_{ 1 }\\ { a }_{ 5 }=2{ a }_{ 4 }-{ a }_{ 3 }=6{ a }_{ 2 }-4{ a }_{ 1 }-2{ a }_{ 2 }+{ a }_{ 1 }=4{ a }_{ 2 }-3{ a }_{ 1 }\\ { a }_{ 6 }=5{ a }_{ 2 }-6{ a }_{ 1 }-3{ a }_{ 1 }+2{ a }_{ 1 }=5{ a }_{ 2 }-4{ a }_{ 1 }\\ { a }_{ 7 }=2{ a }_{ 6 }-{ a }_{ 5 }=10{ a }_{ 2 }-8{ a }_{ 1 }-4{ a }_{ 2 }+3{ a }_{ 1 }=6{ a }_{ 2 }-5{ a }_{ 1 }\\ { a }_{ k }=(k-1){ a }_{ 2 }-(k-2){ a }_{ 1 }\quad \quad k\ge 3\\ \cfrac { { { a }_{ 1 } }^{ 2 }+{ { a }_{ 2 } }^{ 2 }+{ { a }_{ 3 } }^{ 2 }+....+{ { a }_{ 11 } }^{ 2 } }{ 11 } =90$

$={ { a }_{ 1 } }^{ 2 }+{ { a }_{ 2 } }^{ 2 }+\overset { 11 }{ \underset { k=3 }{ \sum } } \left[ { (k-1) }^{ 2 }{ { a }_{ 2 } }^{ 2 }+{ (k-2) }^{ 2 }{ { a }_{ 1 } }^{ 2 }-2(k-1)(k-2){ a }_{ 1 }{ a }_{ 2 } \right] =990\\ ={ { a }_{ 1 } }^{ 2 }+{ { a }_{ 2 } }^{ 2 }+{ { a }_{ 2 } }^{ 2 }\overset { 11 }{ \underset { k=3 }{ \sum } } \left( { k }^{ 2 }+1-2k \right) +\overset { 11 }{ \underset { k=3 }{ \sum } } \left( { k }^{ 2 }+4-4k \right) { { a }_{ 1 } }^{ 2 } \\ -2{ a }_{ 1 }{ a }_{ 2 }\overset { 11 }{ \underset { k=3 }{ \sum } } \left( { k }^{ 2 }-3k+2 \right) =990$

$={ { a }_{ 1 } }^{ 2 }+{ { a }_{ 2 } }^{ 2 }+\overset { 11 }{ \underset { k=3 }{ \sum } } \left( { { a }_{ 2 } }^{ 2 }+{ { a }_{ 1 } }^{ 2 }-2{ a }_{ 1 }{ a }_{ 2 } \right) { k }^{ 2 }+\overset { 11 }{ \underset { k=3 }{ \sum } } k\left( -2{ { a }_{ 2 } }^{ 2 }-4{ { a }_{ 1 } }^{ 2 }+6{ a }_{ 1 }{ a }_{ 2 } \right) \\ +\left( { { a }_{ 2 } }^{ 2 }+4{ { a }_{ 1 } }^{ 2 }-4{ a }_{ 1 }{ a }_{ 2 } \right) \overset { 11 }{ \underset { k=3 }{ \sum } } (1)=990\\ ={ { a }_{ 1 } }^{ 2 }+{ { a }_{ 2 } }^{ 2 }+\left( { { a }_{ 2 } }^{ 2 }+{ { a }_{ 1 } }^{ 2 }-2{ a }_{ 1 }{ a }_{ 2 } \right) { \left[ \cfrac { k(k+1)(2k+1) }{ 6 } \right] }_{ 2 }^{ 11 }+\left( -2{ { a }_{ 2 } }^{ 2 }-4{ { a }_{ 1 } }^{ 2 }+6{ a }_{ 1 }{ a }_{ 2 } \right) { \left[ \cfrac { k(k+1) }{ 2 } \right] }_{ 2 }^{ 11 } \\ +\left( { { a }_{ 2 } }^{ 2 }+4{ { a }_{ 1 } }^{ 2 }-4{ a }_{ 1 }{ a }_{ 2 } \right) { \left[ k \right] }_{ 2 }^{ 11 }=990$

$={ { a }_{ 1 } }^{ 2 }+{ { a }_{ 2 } }^{ 2 }+\left( { { a }_{ 2 } }^{ 2 }+{ { a }_{ 1 } }^{ 2 }-2{ a }_{ 1 }{ a }_{ 2 } \right) \left[ \cfrac { 11(12)(23)-2(3)(5) }{ 6 } \right] \\ +\left( -2{ { a }_{ 2 } }^{ 2 }-4{ { a }_{ 1 } }^{ 2 }+6{ a }_{ 1 }{ a }_{ 2 } \right) \left[ \cfrac { 11(12)-2(3) }{ 2 } \right] +\left( { { a }_{ 2 } }^{ 2 }+4{ { a }_{ 1 } }^{ 2 }-4{ a }_{ 1 }{ a }_{ 2 } \right) \left[ 11-2 \right] =990$

$={ { a }_{ 1 } }^{ 2 }+{ { a }_{ 2 } }^{ 2 }+\left( { { a }_{ 2 } }^{ 2 }+{ { a }_{ 1 } }^{ 2 }-2{ a }_{ 1 }{ a }_{ 2 } \right) (501)+\left( -2{ { a }_{ 2 } }^{ 2 }-4{ { a }_{ 1 } }^{ 2 }+6{ a }_{ 1 }{ a }_{ 2 } \right) 63 \\ +\left( { { a }_{ 2 } }^{ 2 }+4{ { a }_{ 1 } }^{ 2 }-4{ a }_{ 1 }{ a }_{ 2 } \right) 9=990\\ =(1+501+9-126){ { a }_{ 2 } }^{ 2 }+{ { a }_{ 1 } }^{ 2 } (1+501-252+36)+2{ a }_{ 1 }{ a }_{ 2 }(-501+189-18)=990\quad \left( \because { a }_{ 1 }=15 \right) \\ =385{ { a }_{ 2 } }^{ 2 }+64350-9900{ a }_{ 2 }=990\\ =385{ { a }_{ 2 } }^{ 2 }-9900{ a }_{ 2 }+63360=0\\ { a }_{ 2 }=\cfrac { +9900\pm \sqrt { { (9900) }^{ 2 }-4\times 385\times 63360 } }{ 2\times 385 } \\ { a }_{ 2 }=12\\ \cfrac { \left( { a }_{ 1 }+{ a }_{ 2 }+{ a }_{ 3 }+....+{ a }_{ 11 } \right) }{ 11 } \\ =\cfrac { \left[ { a }_{ 1 }+{ a }_{ 2 }+\overset { 11 }{ \underset { k=3 }{ \sum } } (k-1){ a }_{ 2 }-(k-2){ a }_{ 1 } \right] }{ 11 } \\ =\cfrac { \left[ { a }_{ 1 }+{ a }_{ 2 }+({ a }_{ 2 }-{ a }_{ 1 })\overset { 11 }{ \underset { k=3 }{ \sum } } k+(2{ a }_{ 1 }-{ a }_{ 2 })\overset { 11 }{ \underset { k=3 }{ \sum } } (1) \right] }{ 11 } \\ =\cfrac { 15+12+(-3){ \left[ \cfrac { k(k+1) }{ 2 } \right] }_{ 2 }^{ 11 }+(2\times 15-12){ \left[ k \right] }_{ 2 }^{ 11 } }{ 11 } \\ =\cfrac { (27-3\times 63+18\times 9) }{ 11 } \\ =0$

$\sum_{r=1}^{n}(r)(r+1)(2r+3)=an^4+bn^3+cn^2+dn+e$ , then

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$a+c=b+d$
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$e=0$
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$a, b-\dfrac23, c-1$ are in A.P.
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$\dfrac{c}a$ is an integers

Explanation

$\displaystyle \sum _{ r=1 }^{ n }{ r } \left( r+1 \right) \left( 2r+3 \right) =\sum _{ r=1 }^{ n }{ \left( { 2r }^{ 3 }+{ 3r }^{ 2 }+{ 2r }^{ 2 }+3r \right) } =\sum _{ r=1 }^{ n }{ \left( { 2r }^{ 3 }+{ 5r }^{ 2 }+3r \right) }$

$\displaystyle =2{ \left( \frac { n\left( n+1 \right) }{ 2 } \right) }^{ 2 }+5\left( \frac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 } \right) +3\left( \frac { n\left( n+1 \right) }{ 2 } \right)$

$\displaystyle =\frac { 1 }{ 2 } \left( { n }^{ 2 }\left( { n }^{ 2 }+1+2n \right) \right) +\frac { 5 }{ 6 } \left( n\left( { 2n }^{ 2 }+3n+1 \right) \right) +\frac { 3 }{ 2 } \left( { n }^{ 2 }+n \right)$

$\displaystyle =\frac { 1 }{ 2 } \left( { n }^{ 4 }+{ 2n }^{ 3 }+n^2 \right) +\frac { 5 }{ 6 } \left( { 2n }^{ 3 }+{ 3n }^{ 2 }+n \right) +\frac { 3 }{ 2 } \left( { n }^{ 2 }+n \right)$

$\displaystyle =\frac { 1 }{ 2 } { n }^{ 4 }+\frac { 16 }{ 6 } { n }^{ 3 }+\dfrac92{ n }^{ 2 }+\frac { 14 }{ 6 } n$

$\sum_{k=1}^{n}\left ( \sum_{m=1}^{k} m^{2} \right )=an^{4}+bn^{3}+cn^{2}+dn+e$ then

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$a=\dfrac{1}{12}$
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$b=\dfrac{1}{16}$
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$d=\dfrac{1}{6}$
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$e=0$

Explanation

$\sum _{ k=1 }^{ n } \left( \sum _{ m=1 }^{ k } m^{ 2 } \right) =an^{ 4 }+bn^{ 3 }+cn^{ 2 }+dn+e$

$\Rightarrow \sum _{ k=1 }^{ n } \left[ \dfrac { k\left( k+1 \right) \left( 2k+1 \right) }{ 6 } \right] =an^{ 4 }+bn^{ 3 }+cn^{ 2 }+dn+e$

$\Rightarrow \dfrac { 1 }{ 6 } \left\{ 2\sum _{ k=1 }^{ n }{ { k }^{ 3 } } +3\sum _{ k=1 }^{ n }{ { k }^{ 2 } } +\sum _{ k=1 }^{ n } k \right\} =an^{ 4 }+bn^{ 3 }+cn^{ 2 }+dn+e$

$\Rightarrow \dfrac { 1 }{ 6 } \left\{ \dfrac { { k }^{ 2 }{ \left( k+1 \right) }^{ 2 } }{ 2 } +\dfrac { { k }{ \left( k+1 \right) }{ \left( 2k+1 \right) } }{ 2 } +\dfrac { { k }{ \left( k+1 \right) } }{ 2 } \right\} =an^{ 4 }+bn^{ 3 }+cn^{ 2 }+dn+e$

$\Rightarrow \dfrac { n^{ 4 } }{ 12 } +\dfrac { n^{ 3 } }{ 3 } +\dfrac { { 5n }^{ 2 } }{ 12 } +\dfrac { { n } }{ 6 } =an^{ 4 }+bn^{ 3 }+cn^{ 2 }+dn+e$

Therefore,

$a=\dfrac { 1 }{ 12 }$ ;

$b=\dfrac { 1 }{ 3 }$ ;

$c=\dfrac { 5 }{ 12 }$ ;

$d=\dfrac { 1 }{ 6 }$ ;

$e=0$

Ans: A,C,D

Evaluate

$\sum_{r=1}^{n}\left [ \sum_{k=1}^{r}k \right ] \left [ \log_{1/2}\sqrt{(4x-x^{2})} \right ]^{r}$ . Find

$x$ for which summation is a finite number as

$n\rightarrow \infty$

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$x\in \left ( 0,2+\displaystyle \frac{\sqrt{15}}{2} \right )$
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$x\in \left ( 0,2-\displaystyle \frac{\sqrt{15}}{2} \right )$
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$x\in \left ( 0,-2+\displaystyle \frac{\sqrt{15}}{2} \right )$
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$x\in \left ( 0,-2-\displaystyle \frac{\sqrt{15}}{2} \right )$

Explanation

We have,

$\sum_{r=1}^{n}\left [ \sum_{k=1}^{r}k \right ] \left [ \log_{1/2}\sqrt{(4x-x^{2})} \right ]^{r}$

$=\sum_{r=1}^{n} \displaystyle \frac{r(r+1)}{2}\left [ \log_{1/2} \sqrt{(4x-x^{2})}\right ]^{r}$

$=\sum_{r=1}^{n} \displaystyle \frac{r(r+1)}{2}A^{r}$ [where

$A= \log_{1/2} \sqrt{(4x-x^{2})}$ ]

$=\displaystyle \frac{1}{2}\sum_{r=1}^{n}r(r+1)A^{r}$

$=\displaystyle \frac{1}{2}\left [ (1.2)A+(2.3)A^{2}+(3.4)A^{3}+....+n(n+1)A^{n} \right ]$

$S=\displaystyle \frac{1}{2}S_{1}$ (say)

$\cdots (1)$

$\therefore S_{1}=(1.2)A+(2.3)A^{2}+(3.4)A^{3}+....+(n-1)nA^{n-1}+n(n+1)A^{n}$

$\Rightarrow S_{1}=2A+6A^{2}+12A^{3}+.....+(n-1)nA^{n-1}+n(n+1)A^{n}$

$\therefore AS_{1}=0+2A^{2}+6A^{3}+....+(n-1)nA^{n}+n(n+1)A^{n+1}$
Subtracting, we get,

$(1-A)S_{1}=2A+4A^{2}+6A^{3}+....+2nA^{n}-n(n+1)A^{n+1}$

$\Rightarrow (1-A)S_{1}=2[A+2A^{2}+3A^{3}+...+nA^{n}]-n(n+1)A^{n+1}$

$=2S_{2}-n(n+1)A^{n+1}$ (say)

$\cdots (2)$

$\therefore S_{2}=A+2A^{2}+3A^{3}+....+(n-1)A^{n-1}+n A^{n}$
from

$(2)$ ,

$(1-A)S_{1}=\displaystyle \frac{2A(1-A^{n})}{(1-A)^{2}}-\displaystyle \frac{2nA^{n+1}}{(1-A)}-n(n+1)A^{n+1}$

$\therefore S_{1}=\displaystyle \frac{2A(1-A^{n})}{(1-A)^{3}}-\displaystyle \frac{2nA^{n+1}}{(1-A)^{2}}-\displaystyle \frac{n(n+1)A^{n+1}}{(1-A)}$
now from

$(1)$ ,

$S=\displaystyle \frac{1}{2}S_{1}$

$\therefore S=\displaystyle \frac{A(1-A^{n})}{(1-A)^{3}}-\displaystyle \frac{nA^{n+1}}{(1-A)^{2}}-\displaystyle \frac{n(n+1)A^{n+1}}{2(1-A)}$
Now, for

$S$ to be finite we must have

$|A| < 1$

$S_{\infty }=\displaystyle \frac{A(1-0)}{(1-A)^{3}}-0-0$

$=\displaystyle \frac{A}{(1-A)^{3}}=\displaystyle \frac{\log_{1/2}\sqrt{(4x-x^{2})}}{(1-\log_{1/2}\sqrt{(4x-x^{2})})^{3}}=finite$

$(\because A<1)$
i.e

$\log_{1/2}\sqrt{(4x-x^{2})} < 1$

$\Rightarrow \sqrt{4x-x^{2}} > \displaystyle \frac{1}{2}$

$\Rightarrow 4x-x^{2} > \displaystyle \frac{1}{4}$

$\Rightarrow 4x^{2}-16x+1 < 0$

$\therefore 2-\displaystyle \frac{\sqrt{15}}{2} < x < 2+\displaystyle \frac{\sqrt{15}}{2} \cdots (3)$
and

$4x-x^{2} > 0$

$\Rightarrow x^{2}-4x < 0$

$\Rightarrow x(x-4) < 0$

$\therefore 0<x<4 \cdots (4)$
combining

$(3)$ and

$(4)$ we get

$x\in \left ( 0,2+\displaystyle \frac{\sqrt{15}}{2} \right )$

It is known that

$\sum_{r=1}^{\infty }\frac{1}{\left ( 2r-1 \right )^{2}}=\frac{\pi ^{2}}{8}$ . Then

$\sum_{r=1}^{\infty }\frac{1}{r^{2}}$ is equal to

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$\dfrac{\pi ^{2}}{24}$
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$\dfrac{\pi ^{2}}{3}$
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$\dfrac{\pi ^{2}}{6}$
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none of these

Explanation

Given

$\dfrac{1}{1^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{5^{2}}+... = \dfrac{\pi^{2}}{8}$

Now,

$S_{n} = \dfrac{1}{1^{2}}+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{4^{2}}+...$

$\Rightarrow S_{n} = \left\{\dfrac{1}{1^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{5^{2}}+... \right\} + \left\{\dfrac{1}{2^{2}}+\dfrac{1}{4^{2}}+\dfrac{1}{6^{2}}+... \right\}$

$\Rightarrow S_{n} = \dfrac{\pi^{2}}{8} + \dfrac{1}{2^{2}}\left\{\dfrac{1}{1^{2}}+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{4^{2}}+... \right\}$

$S_{n} = \dfrac{\pi^{2}}{8}+\dfrac{1}{4}S_{n}$

$\Rightarrow \dfrac{3}{4}S_{n}= \dfrac{\pi^{2}}{8}$

$\therefore S_{n} = \dfrac{\pi^{2}}{6}$

$\displaystyle \frac{1}{1^2}\, +\, \displaystyle \frac{1}{2^2}\, +\, \displaystyle \frac{1}{3^2}$ + .......... upto

$\infty\, =\, \displaystyle \frac{\pi^2}{6}$ , then

$\displaystyle \frac{1}{1^2}\, +\, \displaystyle \frac{1}{3^2}\, +\, \displaystyle \frac{1}{5^2}$ + ........... = .......... .

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$\pi^2 / 8$
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$\pi^2 / 12$
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$\pi^2 / 3$
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$\pi^2 / 9$

Explanation

$\displaystyle \frac{1}{1^2}\, +\, \displaystyle \frac{1}{2^2}\, +\, \displaystyle \frac{1}{3^2}\, +\, \displaystyle \frac{1}{4^2}\, ..........\, \infty\, =\, \displaystyle \frac{\pi^2}{6}$
Now

$\displaystyle \frac{1}{1^2}\, +\, \displaystyle \frac{1}{3^2}\, +\, \displaystyle \frac{1}{5^2}\, ...........$

$=\, \displaystyle \frac{1}{1^2}\, +\, \displaystyle \frac{1}{2^2}\, +\,\displaystyle \frac{1}{3^2}\,+\, \displaystyle \frac{1}{4^2}\, ..........\,\infty\, -\, \left ( \displaystyle \frac{1}{2^2}\, +\, \displaystyle \frac{1}{4^2}\, +\, \displaystyle \frac{1}{6^2}\, ..........\, \infty \right )$

$=\, \displaystyle \frac{\pi^2}{6}\, -\, \displaystyle \frac{1}{2^2}\, \left ( \displaystyle \frac{1}{1^2}\, +\, \displaystyle \frac{1}{2^2}\, +\, \displaystyle \frac{1}{3^2} \, ..........\, \infty \right )$

$=\, \displaystyle \frac{\pi^2}{6}\, -\, \displaystyle \frac{1}{4}\, .\, \displaystyle \frac{\pi^2}{6}\, =\, \displaystyle \frac{\pi^2}{8}$

Let

$\displaystyle \left \{ a_{n} \right \}\: and\: \left \{ b_{n} \right \}$ are two sequences given by

$\displaystyle a_{n}=\left ( x \right )^{1/2^{n}}+\left ( y \right )^{1/2^{n}}\: \: and\: \: b_{n}=\left ( x \right )^{1/2^{n}}-\left ( y \right )^{1/2^{n}}$ for all n

$\displaystyle \epsilon$ N. The value of

$\displaystyle a_{1}a_{2}a_{3}...a_{n}$ is equal to

0%
x - y
0%
$\displaystyle \frac{x+y}{b_{n}}$
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$\displaystyle \frac{x-y}{b_{n}}$
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$\displaystyle \frac{xy}{b_{n}}$

Explanation

${ a }_{ n }=\cfrac { \left( { x }^{ 1/{ 2 }^{ n } }+{ y }^{ 1/{ 2 }^{ n } } \right) \left( { x }^{ 1/{ 2 }^{ n } }-{ y }^{ 1/{ 2 }^{ n } } \right) }{ \left( { x }^{ 1/{ 2 }^{ n } }-{ y }^{ 1/{ 2 }^{ n } } \right) } \\ { a }_{ n }=\cfrac { \left( { x }^{ 1/{ 2 }^{ n-1 } }-{ y }^{ 1/{ 2 }^{ n-1 } } \right) }{ \left( { x }^{ 1/{ 2 }^{ n } }-{ y }^{ 1/{ 2 }^{ n } } \right) } =\cfrac { { b }_{ n-1 } }{ { b }_{ n } } \\ { a }_{ 1 }{ a }_{ 2 }{ a }_{ 3 }......{ a }_{ n }=\cfrac { { b }_{ 0 } }{ { b }_{ 1 } } \times \cfrac { { b }_{ 1 } }{ { b }_{ 2 } } \times \cfrac { { b }_{ 2 } }{ { b }_{ 3 } } \times .....\times \cfrac { { b }_{ n-1 } }{ { b }_{ n } } \\ { a }_{ 1 }{ a }_{ 2 }{ a }_{ 3 }......{ a }_{ n }=\cfrac { { b }_{ 0 } }{ { b }_{ n } } =\cfrac { (x-y) }{ { b }_{ n } }$

If a number sequence begins

$1, 3, 4, 6, 7, 9, 10, 12 . . .$ , which of the following numbers does NOT appear in the sequence?

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$34$
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$43$
0%
$57$
0%
$65$
0%
$72$

Explanation

$1,3,4,6,7,9,10,12...$

$=\left( 1,4,7,10... \right) ,\left( 3,6,9,12... \right)$
If a number can be written in the form of

$\left( 3k \right)$ or

$\left( 3k+1 \right)$ ., then it will appear in the sequence.
Now check options
For

$34=3\left( 11 \right) +1=3k+1$ , where

$k=11$
For

$43=3\left( 14 \right) +1=3k+1$ , where

$k=14$
For

$57=3\left( 19 \right) =3k$ , where

$k=19$
For

$72=3\left( 24 \right)$ , where

$k=24$
For Here

$65=3\left( 21 \right) +2=3k+2$

$\left( 3k+2 \right)$ will not appear.
Therefore,

$65$ will not appear in the sequence.

Sum of the series

$\displaystyle \sum_{r=1}^{88}\left ( -1 \right )^{r+1}\frac{1}{\sin ^{2}\left ( r+1 \right )^{\circ}-\sin ^{2}1^{\circ}}$ is equal to

0%
$\displaystyle \frac{\cot 2^{\circ}}{\sin 2^{\circ}}$
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$\displaystyle \frac{-\cot 2^{\circ}}{\sin 2^{\circ}}$
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$\displaystyle \cot 2^{\circ}$
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$\displaystyle \frac{\cot 2^{\circ}}{\sin ^{2}2^{\circ}}$

Explanation

$\displaystyle \sum_{88}^{r=1}(-1)^{r+1}\frac{1}{\sin ^{2}(r+1)^{\circ}-\sin ^{2}1^{\circ}}$

$\displaystyle =\left (\frac{1}{\sin 1^{\circ}\sin 3^{\circ}}+\frac{1}{\sin 3^{\circ}\sin 5^{\circ}}+...+\frac{1}{\sin 87^{\circ}\sin 89^{\circ}} \right )-\left (\frac{1}{\sin 2^{\circ}\sin 4^{\circ}}+\frac{1}{\sin 4^{\circ}\sin 6^{\circ}}+.....+\frac{1}{\sin 88^{\circ}\sin 90^{\circ}} \right )$

$\displaystyle = \frac{1}{\sin 2^{\circ}}\left [ \left ( \cot 1^{\circ}-\cot 3^{\circ} \right ) +\left ( \cot 3^{\circ}-\cot 5^{\circ} \right )+....+\left ( \cot 87^{\circ}-\cot 89^{\circ} \right ) \right ]$

$\displaystyle =\left [ \left ( \cot 2^{\circ}-\cot 4^{\circ} \right )+ \left ( \cot 4^{\circ}-\cot 6^{\circ} \right )+........+\left ( \cot 88^{\circ}-\cot 90^{\circ} \right ) \right ]$

$\displaystyle =\frac{\cot 2^{\circ}-0}{\sin 2^{\circ}}=\frac{\cot 2^{\circ}}{\sin 2^{\circ}}$

The expression

$\displaystyle \frac {2^2 + 1} {2^2 - 1} + \frac {3^2 + 1} {3^2 - 1} + \frac {4^2 + 1} {4^2 - 1} + ........... + \frac {(2011)^2 + 1} {(2011)^2 - 1}$
lies in the interval

0%
$\displaystyle (2011, 2010 \frac {1} {2} )$
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$\displaystyle \left( 2011 - \frac {1} {2011} , 2011 - \frac {1} {2012} \right)$
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$\displaystyle (2011, 2011 \frac {1} {2} )$
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$\displaystyle (2012, 2012 \frac {1} {2} )$

Explanation

We can write the expressions as

$\displaystyle \sum _{r = 2}^{2011} {\frac {r^2 + 1}{r^2 - 1}} = \sum \left[1 + \frac {2} {(r + 1) (r - 1)} \right]$

$\displaystyle = \sum \left[ 1 + \frac {1} {r-1} - \frac {1} {r+1} \right]$

Putting

$r = 2, 3, ............, 2011$

$\displaystyle = 2010 + 1 + \frac {1} {2} - \frac {1} {2012} - \frac {1} {2011}$

$\displaystyle = 2011 + \frac {1} {2} - \left[ \frac {1} {2012} + \frac {1} {2011} \right]$

this lies between

$\displaystyle (2011, 2011 \frac {1} {2})$

Numbers can be classified into two categories,depending on their divisible conditions.

They are (i) Even numbers

$(2p) \vee p \epsilon N$ (ii) odd numbers

$(2p + 1) \vee p \epsilon N$

$a_1, a_2 ...... a_{2013}$ are integers, not necessarily distinct.

$x = (-1)^{a_1}+(-1)^{a_2}+.....+(-1)^{a_{1006}}$

$y = (-1)^{a_{1007}}+(-1)^{a_{1008}}+......+(-1)^{a_{2013}}$

Then which of the following is true?

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$(-1)^x = 1; (-1)^y = 1$
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$(-1)^x = 1; (-1)^y = -1$
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$(-1)^x = -1; (-1)^y = 1$
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$(-1)^x = -1; (-1)^y = -1$

If j, k, and n are consecutive integers such that

$0 < j < k < n$ and the units (ones) digit of the product jn is 9, what is the units digit of k ?

0%
0
0%
1
0%
2
0%
3
0%
4

Explanation

There are only a few ways you can make a product have a units digit of $9$ . Either both numbers you’re multiplying have to be $3$ , or one has to be $1$ and one has to be $9$ . Since we’re dealing with consecutive integers, $j$ and $n$ can’t both end in $3$ , so they’re going to have to end in $1$ and $9$ .

It’s important to remember that although we only really care about the units digits in this problem, we’re dealing with numbers that might (or in fact, must) be $2$ or more digits. That’s why you can have $k's$ units digit be $0$ .

Say $j= 19$ , $k= 20$ , and $n= 21$ . Then $jn = (19)(21) = 399$ (units digit is $9$ ), and the units digit of $k = 0$ .

Hence option A is correct.

$\displaystyle\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\cdots\cdots+\frac{1}{n(n+1)}=$

0%
$\displaystyle\frac{1}{n+1}$
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$\displaystyle\frac{n+2}{n(n+1)}$
0%
$\displaystyle\frac{n+3}{n(n+1)}$
0%
$\displaystyle\frac{n}{n+1}$

Explanation

$S_n=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{n(n+1)}=$

The general term of this series can be written as

$T_n=\displaystyle\sum^n_{n=1}\dfrac{1}{n(n+1)}$

$S_n=\displaystyle\sum^n_{n=1}\dfrac{1}{n(n+1)}$

$S_n=\displaystyle\sum^n_{n=1}\dfrac{(n+1)-(n)}{n(n+1)}$

$S_n=\displaystyle\sum^n_{n=1}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)$

on putting the values of n

$S_n=\dfrac{1}{1}-\dfrac{1}{2}$

$+\dfrac{1}{2}-\dfrac{1}{3}$

$+\dfrac{1}{n-1}-\dfrac{1}{n}$

$+\dfrac{1}{n}-\dfrac{1}{n+1}$

$S_n=1-\dfrac{1}{n+1}\\$

$S_n=\dfrac{n+1-1}{n+1}\\$

$S_n=\dfrac{n}{n+1}$ .

It is given that

$\sum_{r = 1}^{\infty} \dfrac{1}{(2 r - 1)^2} = \dfrac{\pi^2}{8}$ , then

$\sum_{r = 1}^{\infty} \dfrac{1}{r^2}$ is equal to

0%
$\dfrac{\pi^2}{24}$
0%
$\dfrac{\pi^2}{3}$
0%
$\dfrac{\pi^2}{6}$
0%
None of these

Explanation

$^{\infty } _{r=1}\Sigma \dfrac{1}{\left ( 2r-1 \right )^{2}}=\dfrac{\pi^{2} }{8}$

$\dfrac{1}{1^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{5^{2}}+\dfrac{1}{7^{2}}+........=\dfrac{\pi^{2} }{8}$

$I= \dfrac{1}{1^{2}}+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{4^{2}}+\dfrac{1}{5}+\dfrac{1}{6^{2}}+\dfrac{1}{7^{2}}$

$I=\left ( \dfrac{1}{2^{2}}+\dfrac{1}{4^{2}}+\dfrac{1}{6^{2}}....... \right )+\left ( \dfrac{1}{1^{2}}+\dfrac{1}{3^{2}}+........ \right )$

$=\dfrac{1}{2^{2}}\left ( \dfrac{1}{1^{2}}+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}.......... \right )+\dfrac{\pi^{2} }{8}$

$\dfrac{4I-I}{4}=\dfrac{\pi^{2} }{8}$

$\dfrac{3I}{4}=\dfrac{\pi ^{2}}{8}$

$I=\dfrac{\pi ^{2}}{6}$

When

$x< 1$ , find the sum of the infinite series

$\cfrac { 1 }{ \left( 1-x \right) \left( 1-{ x }^{ 3 } \right) } +\cfrac { { x }^{ 2 } }{ \left( 1-{ x }^{ 3 } \right) \left( 1-{ x }^{ 5 } \right) } +\cfrac { { x }^{ 4 } }{ \left( 1-{ x }^{ 5 } \right) \left( 1-{ x }^{ 7 } \right) } +.....\quad$

0%
$\cfrac { 1 }{ x(1-x^2)\left( 1-{ x }^{ 3 } \right) }$
0%
$\cfrac { 1 }{ (1-x)\left( 1-{ x }^{ 2 } \right) }$
0%
$\cfrac { 1 }{ x(1-x)\left( 1-{ x }^{ 2 } \right) }$
0%
$\cfrac { x }{ x^2(1-x)\left( 1-{ x }^{ 3 } \right) }$

The sum of the series

$\sum _{ n=1 }^{ \infty }{ \sin { \left( \cfrac { n!\pi }{ 720 } \right) } }$ is

0%
$\sin { \left( \cfrac { \pi }{ 180 } \right) } +\sin { \left( \cfrac { \pi }{ 360 } \right) } +\sin { \left( \cfrac { \pi }{ 540 } \right) }$
0%
$\sin { \left( \cfrac { \pi }{ 6 } \right) } +\sin { \left( \cfrac { \pi }{ 30 } \right) } +\sin { \left( \cfrac { \pi }{ 120 } \right) } +\sin { \left( \cfrac { \pi }{ 360 } \right) }$
0%
$\sin { \left( \cfrac { \pi }{ 6 } \right) } +\sin { \left( \cfrac { \pi }{ 30 } \right) } +\sin { \left( \cfrac { \pi }{ 120 } \right) } +\sin { \left( \cfrac { \pi }{ 360 } \right) } +\sin { \left( \cfrac { \pi }{ 720 } \right) }$
0%
$\sin { \left( \cfrac { \pi }{ 180 } \right) } +\sin { \left( \cfrac { \pi }{ 360 } \right) }$

Explanation

$=\sum_{n=1}^{\infty} sin(\dfrac{n!\pi}{6!})$

$=sin(\dfrac{1!\pi}{6!})+sin(\dfrac{2!\pi}{6!})+sin(\dfrac{3!\pi}{6!})+...+sin(\dfrac{6!\pi}{6!})+sin(\dfrac{7!\pi}{6!})+....$

$=sin(\dfrac{\pi}{720})+sin(\dfrac{\pi}{360})+sin(\dfrac{\pi}{120})+...+sin(\pi)+sin(7\pi)+....$

$=sin(\dfrac{\pi}{720})+sin(\dfrac{\pi}{360})+sin(\dfrac{\pi}{120})+sin(\dfrac{\pi}{30})+sin(\dfrac{\pi}{6})$ [

$\because sin(n\pi)=0$ ]

The value of

$1000\left[\dfrac {1}{1\times 2}+\dfrac {1}{2\times 3}+\dfrac {1}{3\times 4}+...+\dfrac {1}{999\times 1000}\right]$ is equal to

0%
$1000$
0%
$999$
0%
$1001$
0%
$\dfrac{1}{999}$

Explanation

$=1000[\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{999\times1000}]$

$=1000[\dfrac{2-1}{1\times2}+\dfrac{3-2}{2\times3}+\dfrac{4-3}{3\times4}+...+\dfrac{1000-999}{999\times1000}]$

$=1000[\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{999}-\dfrac{1}{1000}]$ ,

$=1000[1-\dfrac{1}{1000}]$ Only end terms remain, as all intermediate terms cancel out.

$=1000.\dfrac{999}{1000}$

$=999$

Let

$\{a_n\}$ be a sequence of numbers satisfying the relation

$(3-a_{n+1})(6+a_n)=18$ for all

$n\ge 0$ and

$a_0=3$ . Then

$\displaystyle \underset{n\rightarrow \infty}{lim}\dfrac{1}{2^{n+2}}\sum_{j=0}^{n}\dfrac{1}{a_j}$

0%
$\dfrac{1}{18}$
0%
$\dfrac{1}{6}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{3}$

Explanation

Given

$\left( 3-{ a }_{ n+1 } \right) \left( 6+{ a }_{ n } \right) =18$

$3-{ a }_{ n+1 }= \cfrac { 18 }{ 6+{ a }_{ n } }$

${ a }_{ n+1 }=3-\cfrac { 18 }{ 6+{ a }_{ n } } = \cfrac { 3{ a }_{ n } }{ 6+{ a }_{ n } }$

$a_0=3,a_1=1,a_2=\cfrac{3}{7},a_3=\cfrac{1}{5},a_4=\cfrac{3}{31},a_5=\cfrac{1}{21}$

Let

$S={ \sum }_{ j=0 }^{ n }\cfrac { 1 }{ { a }_{ j } } =\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 1 } +\cfrac { 7 }{ 3 } +\cfrac { 8 }{ 1 } +\cfrac { 31 }{ 3 } +\cfrac { 21 }{ 1 } +...\\ \quad S=\cfrac { 1 }{ 3 } +1+\cfrac { 7 }{ 3 } +\cfrac { 5 }{ 1 } +\cfrac { 31 }{ 3 } +\cfrac { 21 }{ 1 } +...+{ a' }_{ n }+{ a' }_{ n+1 }\\ -\underline { \;S=\;\;\;\quad \cfrac { 1 }{ 3 } +1+\cfrac { 7 }{ 3 } +\cfrac { 5 }{ 1 } +\cfrac { 31 }{ 3 } +...+{ a' }_{ n }+{ a' }_{ n+1 } } \\ \quad\; 0=\;\cfrac { 1 }{ 3 } +\cfrac { 2 }{ 3 } +\cfrac { 4 }{ 3 } +\cfrac { 8 }{ 3 } +\cfrac { 16 }{ 3 } +\cfrac { 32 }{ 3 } +....-{ a' }_{ n+1 }\\ { a' }_{ n+1 }=\cfrac { 1 }{ 3 } \left( 1+2+4+8+16+32+.... \right)$

This is a GP with

$r=2 \quad S_n=\cfrac{a(r^n-1)}{(r-1)}$

${ a }_{ n }=\cfrac { 1 }{ 3 } \cfrac { (1){ (2) }^{ n+1 }-1 }{ (2-1) } =\cfrac { 1 }{ 3 } ({ 2 }^{ n+1 }-1)$

${ \sum }_{ i=0 }^{ n }{ a' }_{ n }$

$=\cfrac { 1 }{ 3 } \left[ { \Sigma }_{ i=0 }^{ n }{ 2 }^{ n+1 }-{ \Sigma }_{ i=0 }^{ n }n \right] \\ =\cfrac { 1 }{ 3 } \left[ 2\cfrac { ({ 2 }^{ n }-1) }{ 2-1 } -n \right] \\ =\cfrac { 1 }{ 3 } \left( { 2 }^{ n+1 }-2-n \right)$

$\underset { n\rightarrow \infty }{ \lim } \cfrac { 1 }{ 2n-12 } { \sum }_{ i=0 }^{ n }\cfrac { 1 }{ { a }_{ i } } =\underset { n\rightarrow \infty }{ \lim } \cfrac { 1 }{ 6 } \left[ \cfrac { { 2 }^{ n+1 }-2-n }{ { 2 }^{ n+1 } } \right] \\ =\cfrac { 1 }{ 6 } \underset { n\rightarrow \infty }{ \lim } \left[ 1-\cfrac { 2 }{ { 2 }^{ n+1 } } -\cfrac { n }{ { 2 }^{ n+1 } } \right] =\cfrac { 1 }{ 6 }$

Let

$A=16{-4}+2^{-4}+3^{-4}+4^{-4}+...$ and

$B=1^{-4}+3^{-4}+5^{-4}+7^{-4}+...$ . The ratio

$\dfrac{A}{B}$ in the lowest form is

0%
$\dfrac{16}{15}$
0%
$\dfrac{15}{14}$
0%
$\dfrac{15}{16}$
0%
$\dfrac{13}{12}$

Explanation

$A=16-4+{ 2 }^{ -4 }+{ 3 }^{ -4 }+{ 4 }^{ -4 }+....$

$A=\left( 11+{ 1 }^{ -4 }+{ 2 }^{ -4 }+.... \right)$

$\displaystyle \sum _{ n=1 }^{ \infty }{ \cfrac { 1 }{ { n }^{ 4 } } } =\cfrac { {\pi}^{ 4 } }{ 90 }$

$A=11+\cfrac { { \pi }^{ 4 } }{ 90 }$

$B={ 1 }^{ -4 }+{ 3 }^{ -4 }+{ 5 }^{ -4 }+....$

$=\left( { 1 }^{ -4 }+{ 2 }^{ -4 }+{ 3 }^{ -4 }+....\infty \right) -\left( { 2 }^{ -4 }+{ 4 }^{ -4 }+{ 6 }^{ -4 }+.... \right)$

$B=\displaystyle \cfrac { { \pi }^{ 4 } }{ 90 } -\sum _{ n=1 }^{ \infty }{ \cfrac { 1 }{ { \left( 2n \right) }^{ 4 } } }$

$=\displaystyle \cfrac { { \pi }^{ 4 } }{ 90 } -\cfrac { 1 }{ 16 } \sum _{ n=1 }^{ \infty }{ \cfrac { 1 }{ { n }^{ 4 } } } =\cfrac { { \pi }^{ 4 } }{ 90 } -\cfrac { { \pi }^{ 4 } }{ 90\times 16 }$

$=\left( \cfrac { 15 }{ 16 } \right) \cfrac { { \pi }^{ 4 } }{ 90 }$

$\cfrac {A}{B} =\cfrac { 11+\cfrac { { \pi }^{ 4 } }{ 90 } }{ \left( \cfrac { 15 }{ 16 } \right) \cfrac { { \pi }^{ 4 } }{ 90 } } =\left( \cfrac { 16 }{ 15 } \right) \left[ \cfrac { 90\times 11 }{ { \pi }^{ 4 } } +1 \right]$

Lowest form will be

$\cfrac { 16 }{ 15 }$

The variance of the series

$a,a+d,a+2d,.....a+(2n-1)d,a+2nd$ is

0%
$\cfrac { n(n+1) }{ 2 } { d }^{ 2 }$
0%
$\cfrac { n(n-1) }{ 6 } { d }^{ 2 }$
0%
$\cfrac { n(n+1) }{ 6 } { d }^{ 2 }$
0%
$\cfrac { n(n+1) }{ 3 } { d }^{ 2 }$

Explanation

The mean of the given series

$a+nd$

$\therefore$ Variance

$=\cfrac { 1 }{ 2n+1 } \sum _{ r=0 }^{ 2n }{ { \left\{ (a+rd)-(a+nd) \right\} }^{ 2 } }$

$\Rightarrow$ Variance

$\quad =\cfrac { { d }^{ 2 } }{ 2n+1 } \sum _{ r=0 }^{ 2n }{ { \left| r-n \right| }^{ 2 } }$

$\Rightarrow$ Variance

$=\cfrac { { d }^{ 2 } }{ 2n+1 } \sum _{ r=0 }^{ 2n }{ { (n-r) }^{ 2 } }$

$\Rightarrow$ Variance

$=\cfrac { { 2d }^{ 2 } }{ 2n+1 } \sum _{ r=1 }^{ n }{ { r }^{ 2 } }$

$\Rightarrow$ Variance

$=\cfrac { { 2d }^{ 2 } }{ 2n+1 } \times \cfrac { n(n+1)(2n+1) }{ 6 } \quad =\cfrac { n(n+1) }{ 3 } { d }^{ 2 }$

Which of the following options will continue the given series?

$29$ ,

$34$ ,

$32$ ,

$37$ ,

$35$ , ?

0%
$36$
0%
$39$
0%
$40$
0%
$42$

The value of

$S=\sqrt { 1+\cfrac { 1 }{ { 1 }^{ 2 } } +\cfrac { 1 }{ { 2 }^{ 2 } } } +\sqrt { 1++\cfrac { 1 }{ { 2 }^{ 2 } } +\cfrac { 1 }{ { 3 }^{ 2 } } } +...+\sqrt { 1++\cfrac { 1 }{ { (2014) }^{ 2 } } +\cfrac { 1 }{ { (2015) }^{ 2 } } }$ is

0%
$2015$
0%
$2015-\cfrac { 1 }{ 2015 }$
0%
$2016-\cfrac { 1 }{ 2016 }$
0%
$2014-\cfrac { 1 }{ 2014 }$

Explanation

The first term

$\sqrt{1+\dfrac1{1^2}+\dfrac1{2^2}}=\sqrt{1+1+\dfrac14}=\sqrt{\dfrac94}=\dfrac32=1+\dfrac12=1+1-\dfrac12$

The second term

$\sqrt{1+\dfrac1{2^2}+\dfrac1{3^2}}=\sqrt{1+\dfrac14+\dfrac19}=\sqrt{\dfrac{49}{36}}=\dfrac76=1+\dfrac16=1+\dfrac12-\dfrac13$

The Third term similarily can be written as

$1+\dfrac13-\dfrac14$

Hence,

$S=\left (1+1-\dfrac12\right)+\left (1+\dfrac12-\dfrac13\right)+...+\left (1+\dfrac1{2014}-\dfrac1{2015}\right)=2015-\dfrac1{2015}$

$s_n = \displaystyle \sum_{r=0}^n \dfrac {1}{^nC_r}$ and

$t_n = \displaystyle \sum_{r=0}^n \dfrac {r}{^nC_r} ,$ then

$\dfrac {t_n}{s_n}$ is equal to -

0%
$\dfrac {n}{2}$
0%
$\dfrac {n}{2} -1$
0%
$n - 1$
0%
$\dfrac{2n - 1}{2}$

Explanation

${ S }_{ n }={ \sum }_{ r=0 }^{ n }\cfrac { 1 }{ \; ^{ n }{ C }_{ r } } =\cfrac { 1 }{ \; ^{ n }{ C }_{ 0 } } +\cfrac { 1 }{ \; ^{ n }{ C }_{ 1 } } +\cfrac { 1 }{ \; ^{ n }{ C }_{ 2 } } +...+\cfrac { 1 }{ \; ^{ n }{ C }_{ n-1 } } +\cfrac { 1 }{ \; ^{ n }{ C }_{ n } } \\ { S }_{ n }=1+\left\{ \cfrac { 1 }{ \; ^{ n }{ C }_{ 1 } } +\cfrac { 1 }{ \; ^{ n }{ C }_{ 2 } } +...+\cfrac { 1 }{ \; ^{ n }{ C }_{ n-1 } } \right\} +1\\ \; ^{ n }{ C }_{ 0 }=1,\quad \; ^{ n }{ C }_{ n }=1,\quad \; ^{ n }{ C }_{ r }=\; ^{ n }{ C }_{ n-r }\\ { S }_{ n }=2+2\left\{ \cfrac { 1 }{ \; ^{ n }{ C }_{ 1 } } +\cfrac { 1 }{ \; ^{ n }{ C }_{ 2 } } +...+\cfrac { 1 }{ \; ^{ n }{ C }_{ n/2 } } \right\} \\ \cfrac { { S }_{ n }-2 }{ 2 } =\left\{ \cfrac { 1 }{ \; ^{ n }{ C }_{ 1 } } +\cfrac { 1 }{ \; ^{ n }{ C }_{ 2 } } +...+\cfrac { 1 }{ \; ^{ n }{ C }_{ n/2 } } \right\} \quad \quad \quad (1)\\ { t }_{ n }={ \sum }_{ r=0 }^{ n }\cfrac { r }{ \; ^{ n }{ C }_{ r } } =0+\cfrac { 1 }{ \; ^{ n }{ C }_{ 1 } } +\cfrac { 2 }{ \; ^{ n }{ C }_{ 2 } } +\cfrac { 3 }{ \; ^{ n }{ C }_{ 3 } } +...+\cfrac { n-2 }{ \; ^{ n }{ C }_{ n-2 } } +\cfrac { n-1 }{ \; ^{ n }{ C }_{ n-1 } } +\cfrac { n }{ \; ^{ n }{ C }_{ n } } \\ { t }_{ n }=\cfrac { 1+n-1 }{ \; ^{ n }{ C }_{ 1 } } +\cfrac { 2+n-2 }{ \; ^{ n }{ C }_{ 2 } } +...+\cfrac { n/2+n-n/2 }{ \; ^{ n }{ C }_{ n/2 } } +\cfrac { n }{ 1 } \\ { t }_{ n }=n\left[ \cfrac { 1 }{ \; ^{ n }{ C }_{ 1 } } +\cfrac { 1 }{ \; ^{ n }{ C }_{ 2 } } +...+\cfrac { 1 }{ \; ^{ n }{ C }_{ n/2 } } \right] +n$

From

$(1)$

${ t }_{ n }=n\left( \cfrac { { S }_{ n }-2 }{ 2 } \right) +n\\ { t }_{ n }=n\left( \cfrac { { S }_{ n }-2+2 }{ 2 } \right) =\cfrac { n }{ 2 } { S }_{ n }\\ \cfrac { { t }_{ n } }{ { S }_{ n } } =\cfrac { n }{ 2 }$

Solve the given series:

$\dfrac {1.2^2+2.3^2+3.4^2+...n(n+1)^2}{1.2+2^2.3+3^2.4+...n^2(n+1)}$

0%
$(3n+1)(3n+5)$
0%
$\dfrac {n}{2n+1}$
0%
$\dfrac {3n+1}{3n+5}$
0%
$\dfrac {3n+5}{3n+1}$

Explanation

$=\dfrac{1.2^{2}+2.3^{2}+3.4^{2}+...+n(n+1)^{2}}{1^{2}.2+2^{2}.3+3^{2}.4+...+n^{2}(n+1)}$

$=\dfrac{(2-1).2^{2}+(3-1).3^{2}+(4-1).4^{2}+...+(n+1-1)(n+1)^{2}}{1^{2}.(1+1)+2^{2}.(2+1)+3^{2}.(3+1)+...+n^{2}(n+1)}$

$=\dfrac{[1^{3}+2^{3}+3^{3}+4^{3}+...+(n+1)^{3}]-[1^{2}+2^{2}+3^{2}+4^{2}+...+(n+1)^{2}]}{[1^{3}+2^{3}+3^{3}+...+n^{3}]+[1^{2}+2^{2}+3^{2}+...+n^{2}]}$

$=\dfrac{\left[\frac{(n+1)^{2}(n+2)^{2}}{4}\right]-\left[\frac{(n+1)(n+2)(2n+3)}{6}\right]}{\left[\frac{(n)^{2}(n+1)^{2}}{4}\right]+\left[\frac{(n)(n+1)(2n+1)}{6}\right]}$

$=\dfrac{(n+1)(n+2)[6(n+1)(n+2)-4(2n+3)]}{n(n+1)[6(n)(n+1)+4(2n+1)]}$

$=\dfrac{(n+2)[6n^{2}+10n]}{n[6n^{2}+14n+4]}$

$=\dfrac{2n(n+2)[3n+5]}{2n[3n^{2}+7n+2]}$

$=\dfrac{(n+2)(3n+5)}{(3n+1)(n+2)}$

$=\dfrac{(3n+5)}{(3n+1)}$

$\displaystyle \sum _{ r =1 }^{n}{ t_r = \frac{n(n+1)(n+2)(n+3)}{8} }$ , then

$\displaystyle \sum _{r = 1}^{n}{ \frac{1}{t_r}}$ equals

0%
$\displaystyle - \left( \frac{1}{(n+1)(n+2)} - \frac{1}{2} \right)$
0%
$\displaystyle \left( \frac{1}{(n+1)(n+2)} - \frac{1}{2} \right)$
0%
$\displaystyle \left( \frac{1}{(n+1)(n+2)} + \frac{1}{2} \right)$
0%
$\displaystyle \left( \frac{1}{(n-1)(n-2)} + \frac{1}{2} \right)$

Explanation

Let

${ S }_{ n }=\sum _{ r=1 }^{ n }{ { t }_{ r } }$

ie,

${ t }_{ n }={ S }_{ n }-{ S }_{ n-1 }=\dfrac { n(n+1)(n+2)(n+3) }{ 8 } -\dfrac { (n-1)n(n+1)(n+2) }{ 8 }$

$=\dfrac { n(n+1)(n+2)(n+3-(n-1)) }{ 8 }$

$=\dfrac { n(n+1)(n+2) }{ 2 }$

Let,

$\dfrac { 1 }{ { t }_{ n } } =\dfrac { 2 }{ n(n+1)(n+2) } =\dfrac { A }{ n } +\dfrac { B }{ n+1 } +\dfrac { C }{ n+2 }$

$\Rightarrow \dfrac { A }{ n } +\dfrac { B }{ n+1 } +\dfrac { C }{ n+2 } =\dfrac { A(n+1)+Bn }{ n(n+1) } +\dfrac { C }{ n+2 } =\dfrac { ((A+B)n+A)(n+2)+Cn(n+1) }{ n(n+1)(n+2) }$

$\Rightarrow \dfrac { (A+B+C){ n }^{ 2 }+(3A+2B+C)n+2A }{ n(n+1)(n+2) } =\dfrac { 2 }{ n(n+1)(n+2) }$

ie,

$A=1,B=(-2),C=1$

Therefore,

$\sum _{ r=1 }^{ n }{ \dfrac { 1 }{ { t }_{ r } } =\sum _{ r=1 }^{ n }{ (\dfrac { 1 }{ n } -\dfrac { 2 }{ n+1 } +\dfrac { 1 }{ n+2 } ) } } =\left (\dfrac { 1 }{ 1 } +\boxed { -\dfrac { 2 }{ 2 } \\ +\dfrac { 1 }{ 2 } } +\boxed { +\dfrac { 1 }{ 3 } \\ -\dfrac { 2 }{ 3 } +\\ +\dfrac { 1 }{ 3 } } +\boxed { +\dfrac { 1 }{ 4 } \\ -\dfrac { 2 }{ 4 } \\ +\dfrac { 1 }{ 4 } } +\boxed { ... } +\boxed { +\dfrac { 1 }{ n } \\ -\dfrac { 2 }{ n } \\ +\dfrac { 1 }{ n } } +\boxed { +\dfrac { 1 }{ n+1 } \\ -\dfrac { 2 }{ n+1 } } +\dfrac { 1 }{ n+2 }\right )$

$\Rightarrow \sum _{ r=1 }^{ n }{ \dfrac { 1 }{ { t }_{ r } } } =\dfrac { 1 }{ 2 } -\dfrac { 1 }{ n+1 } +\dfrac { 1 }{ n+2 } =\dfrac { 1 }{ 2 } -\dfrac { 1 }{ (n+1)(n+2) }$

$=-\left(\dfrac {1}{(n+1)(n+2)}-\dfrac {1}{2}\right)$

So, option A is correct

The value of

$\sum _{ n=1 }^{ 9999 }{ \cfrac { 1 }{ \left( \sqrt { n } +\sqrt { n+1 } \right) \left( \sqrt [ 4 ]{ n } +\sqrt [ 4 ]{ n+1 } \right) } }$ is

0%
$9$
0%
$99$
0%
$999$
0%
$9999$

Explanation

Let

${ T }_{ n }=\cfrac { 1 }{ \left( \sqrt { n } +\sqrt { n+1 } \right) \left( \sqrt [ 4 ]{ n } +\sqrt [ 4 ]{ n+1 } \right) } \times \cfrac { \sqrt { n+1 } -\sqrt { n } }{ \sqrt { n+1 } -\sqrt { n } } \\ { T }_{ n }=\cfrac { \sqrt { n+1 } -\sqrt { n } }{ \sqrt [ 4 ]{ n+1 } +\sqrt [ 4 ]{ n } } =\cfrac { \left( \sqrt [ 4 ]{ n+1 } -\sqrt [ 4 ]{ n } \right) \left( \sqrt [ 4 ]{ n+1 } +\sqrt [ 4 ]{ n } \right) }{ \left( \sqrt [ 4 ]{ n+1 } +\sqrt [ 4 ]{ n } \right) }$

${ T }_{ n }=\sqrt [ 4 ]{ n+1 } -\sqrt [ 4 ]{ n } \\ { T }_{ 1 }=\sqrt [ 4 ]{ 2 } -\sqrt [ 4 ]{ 1 } \\ { T }_{ 2 }=\sqrt [ 4 ]{ 3 } -\sqrt [ 4 ]{ 2 } \\ { T }_{ 3 }=\sqrt [ 4 ]{ 4 } -\sqrt [ 4 ]{ 3 }$

${ T }_{ n }=\sqrt [ 4 ]{ n+1 } -\sqrt [ 4 ]{ n }$

Adding all

${ S }_{ n }=\sqrt [ 4 ]{ n+1 } -\sqrt [ 4 ]{ 1 } =\sqrt [ 4 ]{ n+1 } -1$

${ S }_{ 9999 }$

$=\sqrt [ 4 ]{ 9999+1 } -1\\ =\sqrt [ 4 ]{ 10000 } -1\\ =10-1\\ =9$

Let

$(1 + x)^{m} = C_{0} + C_{1}x + C_{2}x^{2} + C_{3}x^{3} + .... + C_{m}x^{m}$ , where

$C_{r} = ^{m}C_{r}$ and

$A = C_{1}C_{3} + C_{2}C_{4} + C_{3}C_{5} + C_{4}C_{6} + .... + C_{m - 2}C_{m}$ , then

0%
$A\geq ^{2m}C_{m - 2}$
0%
$A < ^{2m}C_{m - 2}$
0%
$A < C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + .... C_{m}^{2}$
0%
$A > C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + ....C_{m}^{2}$

In a certain code language, DIPLOMA is written as FERHQIC, then what is the code for PENCILS in the language?

0%
RAPYHKU
0%
RAPYKHU
0%
RPAYKHU
0%
RAPKYHU

$x\in R$ and

$S=1-{ C }_{ 1 }\cfrac { 1+x }{ { \left( 1+nx \right) }^{ } } +{ C }_{ 2 }\cfrac { 1+2x }{ { \left( 1+nx \right) }^{ 2 } } -{ C }_{ 3 }\cfrac { 1+3x }{ { \left( 1+nx \right) }^{ 3 } } +...upto\quad (n+1)$ terms, then

$S$

0%
equal ${x}^{2}$
0%
equals $1$
0%
equals $0$
0%
is independent

The sum to infinite of the series

$S=1+\cfrac { 2 }{ 3 } +\cfrac { 6 }{ { 3 }^{ 2 } } +\cfrac { 6 }{ { 3 }^{ 3 } } +\cfrac { 6 }{ { 3 }^{ 4 } } +.....\quad$ is

0%
$4$
0%
$3$
0%
$2$
0%
$6$

The

$(n+1)^{th}$ term from the end in

$(x - \frac{1}{x})^{3n}$ is

0%
$3nc_n.X^{-n}$
0%
$(-1)^n. 3nc_n .X^{-n}$
0%
$3nc_nX^n$
0%
$(-1)^n. 3nc_n.X^n$

If sum of the series

$\displaystyle \sum_{n = 0}^{\infty} r^{n} = S$ , for

$|r| < 1$ , then sum of the series

$\displaystyle \sum_{n = 0}^{\infty} r^{2n}$

0%
$S^{2}$
0%
$\dfrac {S^{2}}{2S + 1}$
0%
$\dfrac {S^{2}}{S^{2} - 1}$
0%
None of these

The value of

$\displaystyle \sum _{ n=0 }^{ 1947 }{ \cfrac { 1 }{ { 2 }^{ n }+\sqrt { { 2 }^{ 1947 } } } }$ is equal to

0%
$\cfrac { 487 }{ \sqrt { { 2 }^{ 1945 } } }$
0%
$\cfrac { 1946 }{ \sqrt { { 2 }^{ 1947 } } }$
0%
$\cfrac { 1947 }{ \sqrt { { 2 }^{ 1947 } } }$
0%
$\cfrac { 1948 }{ \sqrt { { 2 }^{ 1947 } } }$

Explanation

Let

$S=\sum_{n=0}^{1947}{\dfrac 1{2^n+2^{\frac {1947}2}}}=\sum_{n=0}^{1947}{\dfrac 1{2^{1947-n}+2^{\frac {1947}2}}}$

So,

$2S=\sum_{n=0}^{1947}{\dfrac 1{2^n+2^{\frac {1947}2}}}+\sum_{n=0}^{1947}{\dfrac 1{2^{1947-n}+2^{\frac {1947}2}}}=\sum_{n=0}^{1947}{(\dfrac 1{2^n+2^{\frac {1947}2}}+\dfrac 1{2^{1947-n}+2^{\frac {1947}2}})}$

$\Rightarrow 2S=\sum_{n=0}^{1947}{(\dfrac 1{2^n}(\dfrac 1{1+2^{\frac {1947}2-n}})+(\dfrac 1{2^{\frac {1947}2-n}+1})\dfrac 1{2^{\frac {1947}2}})}=\sum_{n=0}^{1947}{((\dfrac 1{2^{\frac {1947}2-n}+1})(\dfrac 1{2^n}+\dfrac 1{2^{\frac {1947}2}}))}$

$\Rightarrow 2S=\sum_{n=0}^{1947}{((\dfrac 1{2^{\frac {1947}2-n}+1})(\dfrac {2^{\frac {1947}2-n}+1}{2^{\frac {1947}2}}))}=\sum_{n=0}^{1947}{\dfrac 1{\sqrt {2^{1947}}}}=\dfrac {1948}{\sqrt {2^{1947}}}$

So,

$S=\dfrac {974}{\sqrt {2^{1947}}}=\dfrac {487}{\sqrt {2^{1945}}}$

Thus option

$A$ is correct.

$\displaystyle \sum_{r = 0}^{n}{t^3 \left( \frac{^nC_r}{^nC_{r-1}} \right)^2 }$ is equal to

0%
$\displaystyle \frac{n(n+1)(n+2)^2}{12}$
0%
$\displaystyle \frac{n(n+1)^2 (n+2)}{12}$
0%
$\displaystyle \frac{n(n+1)(n+2)}{12}$
0%
None of these

Explanation

We have

$s=\displaystyle\sum _0^nr^3\Big(\dfrac{^nC_r}{^nC_{r-1}}\Big)^2$

$s=\displaystyle\sum _0^nr^3\Big(\dfrac{n-r+1}{r}\Big)^2$

$s=\displaystyle\sum _0^nr\Big({n-r+1}\Big)^2$

$s=\displaystyle\sum _0^nr(n^2+r^2+1-2nr-2r+2n)$

$s=\displaystyle\sum _0^nr(n^2+2n+1)+r^3-2r^2(n+1)$

$s=\displaystyle\sum _0^nr(n+1)^2+r^3-2r^2(n+1)$

$s=\dfrac{n(n+1)^3}{2}+\dfrac{n^2(n+1)^2}{4}-\dfrac{n(n+1)^2(2n+1)}{3}$

Taking

$n(n+1)^{2}$ common, we get

$s=n(n+1)^{2}\Big(\dfrac{(n+1)}{2}+\dfrac{n}{4}-\dfrac{(2n+1)}{3}\Big)$

$s=\dfrac{n(n+1)^2(n+2)}{12}$

B is the correct answer

$x=\dfrac{1}{5}+\dfrac{1.3}{5.10}+\dfrac{1.3.5}{5.10.15}+.....\infty$ then

$3x^2+6x=$

0%
1
0%
2
0%
3
0%
4

The value of

$\displaystyle\sum^{n}_{i=1}\sum^{i}_{j=1}\sum^j_{k=1}1=220$ , then the value of n equals.

0%
$11$
0%
$12$
0%
$10$
0%
$9$

Let

${ T }_{ r }$ and

${ S }_{ r }$ be the

${ r }^{ th }$ term and sum up to

${ r }^{ th }$ term of a series respectively. If for an odd natural number

$n,{ S }_{ n }=n$ and

${ T }_{ n }=\dfrac { { T }_{ n-1 } }{ { n }^{ 2 } }$ , then

${ T }_{ m }$ (

$m$ being even) is:

0%
$\dfrac { 2 }{ 1+{ m }^{ 2 } }$
0%
$\dfrac { 2{ m }^{ 2 } }{ 1+{ m }^{ 2 } }$
0%
$\dfrac { { \left( m+1 \right) }^{ 2 } }{ 2+{ \left( m+1 \right) }^{ 2 } }$
0%
$\dfrac { 2{ \left( m+1 \right) }^{ 2 } }{ 1+{ \left( m+1 \right) }^{ 2 } }$

Explanation

Given,

$S_n=n$ and

$T_n=\dfrac {T_{n-1}}{n^2},$

$n$ is an odd natural number.

$\therefore S_n - S_{n-2} =(n)-(n-2)= 2\\ \Rightarrow T_n + T_{n-1} = 2\\ \Rightarrow \dfrac {T_{n-1}}{n^2}+ T_{n-1}= 2\\ \Rightarrow \left( \dfrac{1}{n^2} + 1 \right) T_{n-1} = 2\\ \Rightarrow \left( \dfrac{1+n^2}{n^2} \right) T_{n-1} = 2\\ \Rightarrow T_{n-1} = \dfrac{2n^2}{1 + n^2}$

Since,

$n$ is odd

$\Rightarrow (n-1)$ is odd.

Replacing

$(n-1)$ by

$m$ we get,

$T_m = \dfrac{2(m+1)^2}{1 + (m+1)^2}$

Sum of the series

$\displaystyle\sum^n_{r=1}(r^2+1)r!$ is?

0%
$(n+1)!$
0%
$(n+2)!-1$
0%
$n-(n+1)!$
0%
$n-(n+2)!$

$a_{1}, a_{2}, ......., a_n(n > 3)$ are all unequal positive real numbers, and

$E = \dfrac{(1 + a_{1} + a_{1}^{2})(1 + a_{2} + a_{2}^{2})......(1 + a_{n} + a_{n}^{2})}{a_{1}, a_{2}, ......., a_{n}}$
then which of the following best describes E?

0%
$E \leq 2^{n}$
0%
$E \geq 3^{n}$
0%
$E > 3^{n}$
0%
$E > 2^{n}$

Explanation

${ a }_{ 1 },{ a }_{ 2 },{ a }_{ 3 },...{ a }_{ n }$ all are unequal

$n>3\\ E=\cfrac { \left( 1+{ a }_{ 1 }+{ a }_{ 1 }^{ 2 } \right) \left( 1+{ a }_{ 2 }+{ a }_{ 2 }^{ 2 } \right) ...\left( 1+{ a }_{ n }+{ a }_{ n }^{ n } \right) }{ { a }_{ 1 }.{ a }_{ 2 }.{ a }_{ 3 }....{ a }_{ n } } \\ =\left( \cfrac { 1 }{ { a }_{ 1 } } +1+{ a }_{ 1 } \right) \left( 1+\cfrac { 1 }{ { a }_{ 2 } } +{ a }_{ 2 } \right) \left( 1+\cfrac { 1 }{ { a }_{ 3 } } +{ a }_{ 3 } \right) ....\left( 1+\cfrac { 1 }{ { a }_{ n } } +{ a }_{ n } \right) \\ A.M.\ge G.M.\\ \cfrac { \cfrac { 1 }{ { a }_{ 1 } } +{ a }_{ 1 }+1 }{ 3 } \ge \sqrt [ 3 ]{ \cfrac { 1 }{ { a }_{ 1 } } .{ a }_{ 1 }.1 } \\ \cfrac { 1 }{ { a }_{ 1 } } +{ a }_{ 1 }+1\ge 3$

${ a }_{ 1 },{ a }_{ 2 },{ a }_{ 3 },...{ a }_{ n }$ are unequal if

$\cfrac { 1 }{ { a }_{ 1 } } +{ a }_{ 1 }+1=3$ ,then all others must be

$>3\\ E=\left( \cfrac { 1 }{ { a }_{ 1 } } +1+{ a }_{ 1 } \right) \left( 1+\cfrac { 1 }{ { a }_{ 2 } } +{ a }_{ 2 } \right) ....\left( 1+\cfrac { 1 }{ { a }_{ n } } +{ a }_{ n } \right) \\ >3.3.....3(ntimes)\\ E>{ 3 }^{ n }$

The sum of the series

${1 \over 2} + {3 \over 4} + {7 \over 8} + {{15} \over {16}} + .......$ up to low upon to n term is to the

0%
$n - 1 + {2^{ - n}}$
0%
${2^{ - n}} - 1$
0%
$n - 2$
0%
None of these

Explanation

$\cfrac { 1 }{ 2 } +\cfrac { 3 }{ 4 } +\cfrac { 7 }{ 8 } +\cfrac { 15 }{ 16 } +....upto\quad n\quad terms\\ =(1-\cfrac { 1 }{ 2 } )+(1-\cfrac { 1 }{ 4 } )+....+(1-\cfrac { 1 }{ { 2 }^{ n } } )\\ =n-(\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 4 } +....+\cfrac { 1 }{ { 2 }^{ n } } )\\ =n-\cfrac { \cfrac { 1 }{ 2 } \times \left( { \left( \cfrac { 1 }{ 2 } \right) }^{ n }-1 \right) }{ \cfrac { 1 }{ 2 } -1 } \\ =n+{ 2 }^{ -n }-1$

The sume of the series

$1^3 - 2^3 + 3^3 - ........ + 9^3$ =

0%
300
0%
125
0%
425
0%
0

The positive integer n for which

$2 \times {2^2} + 3 \times {2^3} + 4 \times {2^4} + ....... + n \times {2^n} = {2^{^{n + 10}}}$ is______

0%
$510$
0%
$511$
0%
$512$
0%
$513$

Explanation

$2 \times 2^2 + 3 \times 2^3 + 4 \times 2^4 ... n \times 2^n = 2^{n + 10}$

It is

$AGP$ since

$2, 3, 4 ... AP$

$2^3 , 2^3, 2^4 .... GP$

let

$S = 2 \times 2^2 + 3.2^3 + 4.2^4 ... n.2^n$

$2S =$

$2.2^3 + 3.2^4 ... (n - 1). 2^n + n . 2^{n + 1}$

__________________________________________

$-S = 2.2^2 + 2^3 + 2^4 ... + 2^n - n . 2^{n + 1}$

$-S = 2.2^2 + 2^3 \left(\dfrac{2^{n-3} - 1}{2-1} \right) - n. 2^{n + 1}$

$-S = 2.2^2 + 2^3 (2^{n - 3} - 1) - n.2^{n +1}$

$= 2^3 + 2^n - 2^3 - n.2^{n+1}$

$= 2^n - 2n.2^n$

$S = 2^n (2n - 1)$

$2^n . 2^{10} = 2^n (2n - 1)$

$2^{10} = 2n - 1$

$1024 = 2n - 1$

$1025 = 2n$

$n = \dfrac{1025}{2}$

$n = 512.5$

integer

$= 512$

The sum to infinite of the series

$1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + {{14} \over {{3^4}}} + ........$

0%
2
0%
3
0%
4
0%
6

Explanation

$s=1+\cfrac { 2 }{ 3 } +\cfrac { 6 }{ 3^{ 2 } } +\cfrac { 10 }{ 3^{ 3 } } +....\\ s-1=\cfrac { 2 }{ 3 } +\cfrac { 6 }{ 3^{ 2 } } +\cfrac { 10 }{ 3^{ 3 } } +...\\ \cfrac { s-1 }{ 3 } =\cfrac { 2 }{ 3^{ 2 } } +\cfrac { 6 }{ 3^{ 3 } } +\cfrac { 10 }{ 3^{ 4 } } +...\\ (s-1)-(\cfrac { s-1 }{ 3 } )=\cfrac { 2 }{ 3 } +\cfrac { 4 }{ 3^{ 2 } } +\cfrac { 4 }{ 3^{ 3 } } +\cfrac { 4 }{ 3^{ 4 } } +......\\ (s-1)-(\cfrac { s-1 }{ 3 } )=\cfrac { 2 }{ 3 } +\cfrac { 4 }{ 3^{ 2 } } [1+\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 3^{ 2 } } +\cfrac { 1 }{ 3^{ 3 } } +....]\\ \cfrac { 3s-3-s+1 }{ 3 } =\cfrac { 2 }{ 3 } +\cfrac { 4 }{ 3^{ 2 } } (\cfrac { 1 }{ 1 } -\cfrac { 1 }{ 3 } )\\ \cfrac { 2 }{ 3 } (s-1)=2\times \cfrac { 2 }{ 3 } \\ s-1=2\\ s=3$

$\displaystyle \lim _{ x\rightarrow 0^+ }{ x\left( \left[ \dfrac { 1 }{ x } \right] +\left[ \dfrac { 5 }{ x } \right] +\left[ \dfrac { 11 }{ x } \right] +\left[ \dfrac { 19 }{ x } \right] +\left[ \dfrac { 29 }{ x } \right] +.......to\quad n\quad terms \right) }=430$ (where [.] denotes the greatest integer function), then

$n=$

0%
$8$
0%
$9$
0%
$10$
0%
$11$

Explanation

$\lim_{x\rightarrow 0^{+}} x\left( \left[\dfrac{1}{x}\right] + \left[\dfrac{5}{x}\right] + \left[\dfrac{11}{x}\right] + \left[\dfrac{19}{x}\right] + \left[\dfrac{29}{x}\right] + ....... to \quad n \quad terms\right) = 430$

$\implies 1 + 5 + 11 + 19 + 29 + ......... n terms = 430$

$\implies 1 + 5 + 11 + 19 + 29 + .........+ t_n = 430$

$1 + 5 + 11 + 19 + 29 + ......... t_n$

By method of difference,

$t_n = n^2 + n - 1$

$\because S_n = \sum t_n$

$\implies \sum t_n = 430$

$\implies \sum (n^2 + n - 1) = 430$

$\implies \sum n^2 + \sum n - \sum 1 = 430$

$\implies \dfrac{n(n+1)(2n+1)}{6} + \dfrac{n(n+1)}{2} - n = 430$

$\implies \dfrac{n(n+1)}{2} \left[ \dfrac{2n+1}{3} + 1\right] - n = 430$

$\implies \dfrac{n(n+1)}{2} \left[ \dfrac{2n+4}{3} \right] - n = 430$

$\implies \dfrac{n(n+1)(n+2)}{3} - n = 430$

$\implies n(n+1)(n+2) - 3n = 430\times 3$

$\implies n^3 + 3n^2 + 2n - 3n = 1290$

$\implies n (n^2 + 3n - 1 ) = 10\times 129$

$\implies n (n^2 + 3n - 1 ) = 10\times (10^2 + 30 - 1)$

$\therefore n = 10$

Option C is correct.

$\displaystyle\sum _{ n=1 }^{ 2013 }{ \tan { \left( \dfrac { \theta }{ { 2 }^{ n } } \right) } } \sec { \left( \dfrac { \theta }{ { 2 }^{ n-1 } } \right) } =\left( \dfrac { \theta }{ { 2 }^{ a } } \right) -\left( \dfrac { \theta }{ { 2 }^{ b } } \right)$ then

$(b+a)$ equals

0%
$2014$
0%
$2012$
0%
$2013$
0%
$2019$

$\sum^5_{n=1}\dfrac{1}{n(n+1)(n+2)(n+3)}=\dfrac{k}{3}$ , then k is equal to?

0%
$\dfrac{55}{336}$
0%
$\dfrac{17}{105}$
0%
$\dfrac{19}{112}$
0%
$\dfrac{1}{6}$

Explanation

$\\ \sum _{ n=1 }^{ 5 }{ \cfrac { 1 }{ n(n+1)(n+2)(n+3) } =\cfrac { k }{ 3 } } \\ \cfrac { 1 }{ 2 } \sum _{ n=1 }^{ 5 }{ \left( \cfrac { 1 }{ n(n+3) } -\cfrac { 1 }{ (n+1)(n+3) } \right) } =\cfrac { k }{ 3 } \\ \\ \sum _{ n=1 }^{ 5 }{ \left[ \cfrac { 1 }{ 3 } \left( \cfrac { 3 }{ n(n+3) } \right) -\cfrac { 1 }{ (n+1)(n+3) } \right] } =\cfrac { 2k }{ 3 } \\ \sum _{ n=1 }^{ 5 }{ \left( \cfrac { 1 }{ 3n } -\cfrac { 1 }{ n+1 } +\cfrac { 1 }{ n+2 } -\cfrac { 1 }{ 3(n+3) } \right) } =\cfrac { 2k }{ 3 } \\ \cfrac { 1 }{ 3 } \left( 1+\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 5 } \right) -\left( \cfrac { 1 }{ 2 } +\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 5 } +\cfrac { 1 }{ 6 } \right) +\left( \cfrac { 1 }{ 3 } +\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 5 } +\cfrac { 1 }{ 6 } +\cfrac { 1 }{ 7 } \right) -\cfrac { 1 }{ 3 } \left( \cfrac { 1 }{ 4 } +\cfrac { 1 }{ 5 } +\cfrac { 1 }{ 6 } +\cfrac { 1 }{ 7 } +\cfrac { 1 }{ 8 } \right) =\cfrac { 2k }{ 3 } \\ \cfrac { 1 }{ 3 } \left( 1+\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 3 } -\cfrac { 1 }{ 6 } -\cfrac { 1 }{ 7 } -\cfrac { 1 }{ 8 } \right) -\left( \cfrac { 1 }{ 2 } -\cfrac { 1 }{ 7 } \right) =\cfrac { 2k }{ 3 } \\ \cfrac { 1 }{ 3 } \left( \cfrac { 235 }{ 3\times 56 } \right) -\cfrac { 5 }{ 14 } =\cfrac { 2k }{ 3 } \\ \cfrac { 235 }{ 3\times 56 } -\cfrac { 5 }{ 14 } =2k\\ \cfrac { 55 }{ 3\times 56 } =2k\\ k=\cfrac { 55 }{ 336 }$

Sum to

$n$ terms of the series

$\dfrac { 1 }{ 1.2.3.4 } +\dfrac { 1 }{ 2.3.4.5 } +\dfrac { 1 }{ 3.4.5.6 } +..........$ , is

0%
$\dfrac { { n }^{ 3 } }{ 3\left( n+1 \right) \left( n+2 \right) \left( n+3 \right) }$
0%
$\dfrac { { n }^{ 3 }+{ 6n }^{ 2 }-3n }{ 6\left( n+2 \right) \left( n+3 \right) \left( n+4 \right) }$
0%
$\dfrac { 15{ n }^{ 2 }+7n }{ 4n\left( n+1 \right) \left( n+5 \right) }$
0%
$\dfrac { { n }^{ 3 }+6{ n }^{ 2 }+11n }{ 18\left( n+1 \right) \left( n+2 \right) \left( n+3 \right) }$

The sum to infinity of the series:

$\dfrac {3}{{1}^{3}}+\dfrac {5}{{1}^{3}+{2}^{3}}+\dfrac {7}{{1}^{3}+{2}^{3}+{3}^{3}}+..$ is-

0%
5
0%
- 4
0%
6
0%
1

The sum of infinite series

$\dfrac{1.3}{2}+\dfrac{3.5}{2^2}+\dfrac{5.7}{2^3}+\dfrac{7.9}{2^4}+...\infty$ .

0%
$21$
0%
$22$
0%
$23$
0%
None

Explanation

$S=\cfrac { 1.3 }{ 2 } +\cfrac { 3.5 }{ { 2 }^{ 2 } } +\cfrac { 5.7 }{ { 2 }^{ 3 } } +\cfrac { 7.9 }{ { 2 }^{ 4 } } +....\infty \\ \cfrac { S }{ 2 } =\cfrac { 1.3 }{ { 2 }^{ 2 } } +\cfrac { 3.5 }{ { 2 }^{ 3 } } +\cfrac { 5.7 }{ { 2 }^{ 4 } } +....\infty \\ S-\cfrac { S }{ 2 } =\cfrac { 3 }{ 2 } +\cfrac { 12 }{ { 2 }^{ 2 } } +\cfrac { 20 }{ { 2 }^{ 3 } } +\cfrac { 28 }{ { 2 }^{ 4 } } +....\infty \\ \cfrac { S }{ 2 } =\cfrac { 3 }{ 2 } +4\left( \cfrac { 3 }{ { 2 }^{ 2 } } +\cfrac { 5 }{ { 2 }^{ 3 } } +\cfrac { 7 }{ { 2 }^{ 4 } } +....\infty \right) \\ \cfrac { S }{ 2 } =\cfrac { 3 }{ 2 } +3+\cfrac { 5 }{ { 2 } } +\cfrac { 7 }{ { 2 }^{ 2 } } +\cfrac { 9 }{ { 2 }^{ 3 } } +....\infty \\ \cfrac { S }{ 4 } =\cfrac { 3 }{ 4 } +\cfrac { 3 }{ 2 } +\cfrac { 5 }{ { 2 }^{ 2 } } +\cfrac { 7 }{ { 2 }^{ 3 } } +....\infty \\ \cfrac { S }{ 2 } -\cfrac { S }{ 4 } =\cfrac { 15 }{ 4 } +\left( \cfrac { 2 }{ 2 } +\cfrac { 2 }{ { 2 }^{ 2 } } +\cfrac { 2 }{ { 2 }^{ 3 } } +....\infty \right) \\ \cfrac { S }{ 4 } =\cfrac { 15 }{ 4 } +\cfrac { 1 }{ 1-\cfrac { 1 }{ 2 } } \\ S=23$

$\dfrac{x}{0.2} + \dfrac{x}{0.3} + \dfrac{x}{0.6} + \dfrac{x}{0.4} + \dfrac{x}{0.5} = 87$ , then the value of x is equal to

0%
0
0%
4
0%
6
0%
1

CBSE Questions for Class 11 Engineering Maths Sequences And Series Quiz 11 - MCQExams.com

0:0:3

Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Sequences And Series Quiz 11 - MCQExams.com

0:0:3

Practice Class 11 Engineering Maths Quiz Questions and Answers

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