MCQ Questions for Class 11 Engineering Maths Sequences And Series Quiz 6

$6, 10, 18, 34, 66$
The first number in the list above is

$6$ . Determine a rule for finding each successive number in the list.

0%
Add $4$ to the preceding number.
0%
Take $\displaystyle \frac { 1 }{ 2 }$ of the preceding number and then add $7$ to that result.
0%
Double the preceding number and then subtract $2$ from that result.
0%
Subtract $2$ from the preceding number and then double that result.
0%
Triple the preceding number and then subtract $8$ from that result.

Explanation

The series is

$6,10,18,34,66$

In the given series:

$6+6=12-2=10$

$10+10=20-2=18$

$18+18=36-2=34$

$34+34=68-2=66$

Hence in this each successive number is obtained by double the preceding number and then subtract

$2$ from the result.

The value of the sum

$1.2.3+2.3.4+3.4.5+...$ upto n terms =

0%
$\dfrac{1}{6}n^2 (2n^2+1)$
0%
$\dfrac{1}{6}(n^2-1)(2n-1)(2n+3)$
0%
$\dfrac{1}{8}(n^2+1)(n^2+5)$
0%
$\dfrac{1}{4}n(n+1)(n+2)(n+3)$

$m, 2m, 4m, . . .$
The first term in the sequence above is

$m$ , and each term thereafter is equal to twice the previous term. If

$m$ is an integer, which of the following could NOT be the sum of the first four terms of this sequence?

0%
$-26$
0%
$-15$
0%
$45$
0%
$75$
0%
$120$

Explanation

The first term in given Geometric series,

$a_1=m$
The common ratio

$r$

$=\dfrac{4m}{2m}=2$
No. of terms,

$n$

$=4$
Applying sum of GP formula,

$S_n=\dfrac{a(1-r^n)}{1-r}$

$=\dfrac{m(1-2^4)}{1-2}$

$=\dfrac{m(1-16)}{-1}=15m$

$m$ is an integer the sum of the first four terms of this sequence should be in multiple of

$15$ .
Hence, option A is correct which is not a multiple of

$15$ .

Identify the missing integer:

$9, 45,$ ____

$, 1125, 5625...$

0%
$220$
0%
$223$
0%
$224$
0%
$225$

Explanation

This is continuous series multiplied by

$5$ .

$9 \times 5 = 45$

$45 \times 5 = 225$

$225 \times 5 = 1,125$

$1,125 \times 5= 5,625.$
So, the missing integer is

$225$ .

The terms of a sequence are defined by

$a_{n} = 3a_{n - 1} - a_{n - 2}$ for

$n > 2$ . Find the value of

$a_{5}$ given that

$a_{1} =4$ and

$a_{2} = 3$ .

0%
$12$
0%
$23$
0%
$25$
0%
$31$
0%
$36$

Explanation

Given ${ a }_{ n }=3{ a }_{ n-1 }-{ a }_{ n-2 }$ , ${a}_{1}=4$ and ${a}_{2} = 3$
${a}_{3} = 3{a}_{2}-{a}_{1} = 9-4=5$
${a}_{4} = 3{a}_{3}-{a}_{2} = 15-3 = 12$
${a}_{5}=3{a}_{4}-{a}_{3} = 36-5 = 31$

For all numbers a and b, let

$\displaystyle a\bigodot b$ be defined by

$\displaystyle a\bigodot b=ab+a+b$ . Then for the numbers

$x$ ,

$y$ and

$z$ , which of the following is/are true?
I.

$\displaystyle x\bigodot y=y\bigodot x$
II.

$\displaystyle \left( x-1 \right) \bigodot \left( x+1 \right) =\left( x\bigodot x \right) -1$
III.

$\displaystyle x\bigodot \left( y+z \right) =\left( x\bigodot y \right) +\left( x\bigodot z \right)$

0%
I only
0%
II only
0%
III only
0%
I and II only
0%
I, II, and III

Explanation

$For(I)$

$x\odot y=y\odot x$

$x\odot y=xy+x+y\Rightarrow x\odot y=y\odot x$

$y\odot x=yx+y+x$

$For(II)$

$\left( x-1 \right) \odot \left( x+1 \right)$

$=(x-1)(x+1)+x-1+x+1$

$={ x }^{ 2 }-1+2x$

$\left( x\odot x \right) -1=x\times x+x+x-1$

${ x }^{ 2 }+2x-1$

$\therefore \left( x-1 \right) \odot \left( x+1 \right) =\left( x\odot x \right) -1$

$For(III)$

$x\odot \left( y+z \right)$

$=x\left( y+z \right) +x+y+z$

$=xy+xz+x+y+z$

$x\odot y=xy+x+y$

$x\odot z=xz+x+z$

$x\odot y+x\odot z=xy+xz+x+y+x+z$

$=xy+xz+2x+y+z$

$\Rightarrow x\odot \left( y+z \right) \neq x\odot y+x\odot z$

$\therefore (I)\& (II)$ are true.

The sum of

$1$ st

$n$ terms of the series

$\cfrac { { 1 }^{ 2 } }{ 1 } +\cfrac { { 1 }^{ 2 }+{ 2 }^{ 2 } }{ 1+2 } +\cfrac { { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 } }{ 1+2+3 } +........$

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

Explanation

General term of the given series is,

$T_r=\dfrac{1^2+2^2+....+r^2}{1+2+.....+r}=\dfrac{\frac{r(r+1)(2r+1)}{6}}{\frac{r(r+1)}{2}}=\dfrac{2r+1}{3}$

Hence required sum

$\displaystyle =\sum_{r=1}^nT_r=\dfrac{1}{3}\sum_{r=1}^n(2r+1)$

$\displaystyle=\dfrac{1}{3} \left(2\sum_{r=1}^n r+\sum_{r=1}^n 1 \right)=\dfrac{1}{3}\left(2\cdot \dfrac{n(n+1)}{2} +n\right)=\dfrac{1}{3}(n^2+n+n)=\dfrac{n(n+2)}{3}$

$m * n = m+(m-1)+(m-2)+ ...... +(m-n)$ , evaluate

$7 * 5$ .

0%
$25$
0%
$20$
0%
$27$
0%
$18$

Explanation

$\Rightarrow$

$m * n=m+(m-1)+(m-2)+....+(m-n)$

$\Rightarrow$ In

$7 * 5$ ,

$m=7$ and

$n=5$

$\Rightarrow$

$7 * 5=7+(7-1)+(7-2)+(7-3)+(7-4)+(7-5)$

$\Rightarrow$

$7 * 5=7+6+5+4+3+2$

$\therefore$

$7 * 5=27$

$N=a^2 + b^2$ is a three-digit number which is divisible bya = 10x + y and b = 10x + z, where z is a prime number, and x and y are natural numbers. If a + b = 31, find the value of N.

0%
565
0%
485
0%
505
0%
485 or 505

$a\odot b = 6\times a - 3\times b$ , evaluate

$(5\odot 3) \odot 20$

0%
$2$
0%
$41$
0%
$66$
0%
$1$

Explanation

$\Rightarrow$

$a\odot b=6\times a-3\times b$ [ Given ]

$\Rightarrow$

$(5\odot 3)\odot 20$

$\Rightarrow$ Let

$a=5$ and

$b=3$

$\Rightarrow$

$(6\times 5-3\times 3)\odot 20$

$\Rightarrow$

$(30-9)\odot 20$

$\Rightarrow$

$21\odot 20$

$\Rightarrow$ Let

$a=21$ and

$b=20$

$\Rightarrow$

$6\times 21-3\times 20$

$\Rightarrow$

$126-60$

$\Rightarrow$

$66$

$\therefore$

$(5\odot 3)\odot 20=66$

The sum of the series,

$\displaystyle \frac{1}{2.3}\cdot 2+ \frac{2}{3.4}\cdot 2^2+\frac{3}{4.5}\cdot 2^3+ ......$ to n terms is _____.

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

Explanation

Let

$S = \displaystyle \frac{1}{2.3}\cdot 2+ \frac{2}{3.4}\cdot 2^2+\frac{3}{4.5}\cdot 2^3+ ......$

General term is given by,

$T_r = \dfrac{r}{(r+1)(r+2)}\cdot 2^r=\dfrac{r\cdot 2^r}{(r+1)(r+2)}=\dfrac{2(r+1)-(r+2)}{(r+1)(r+2)}\cdot 2^r=\dfrac{(r+1)2^{r+1}-(r+2)2^r}{(r+1)(r+2)}$

$\Rightarrow T_r = \dfrac{2^{r+1}}{r+2}-\dfrac{2^r}{r+1}$ , split the terms

Therefore the required sum

$=\displaystyle \sum_{r=1}^nT_r=T_1+T_2+T_3+...........+T_n$

$\quad =\left( \dfrac{2^2}{3}-\dfrac{2}{2}\right)+\left( \dfrac{2^3}{4}-\dfrac{2^2}{3}\right)+\left(\dfrac{2^4}{5}-\dfrac{2^3}{4} \right)+..............+\left(\dfrac{2^{n+1}}{n+2} -\dfrac{2^n}{n+1}\right)$

$=-\dfrac{2}{2}+\dfrac{2^{n+1}}{n+2}=\dfrac{2^{n+1}}{n+2}-1$ , all the remaining terms will get cancelled

The value of the infinite series

$\cfrac{1^2+2^2}{\underline{|3}} + \cfrac{1^2+2^2+3^2}{\underline{|4}}+\cfrac{1^2+2^2+3^3+4^2}{\underline{|5}}........$ is

0%
$e$
0%
$5e$
0%
$\cfrac{5e}{6} -\cfrac{1}{2}$
0%
$\cfrac{5e}{6}$

Explanation

${ T }_{ n }=\cfrac { n(n+1)(2n+1) }{ 6(n+1)! } \\ S={ \sum }_{ i=2 }^{ \infty }\cfrac { n(n+1)(2n+1) }{ 6(n+1)! } ={ \sum }_{ i=2 }^{ \infty }\cfrac { (2n+1) }{ 6(n-1)! } \\ S={ \sum }_{ i=2 }^{ \infty }\cfrac { 2n-2+3 }{ 6(n-1)! } \\ S=\cfrac { 1 }{ 6 } { \sum }_{ i=2 }^{ \infty }\cfrac { 2(n-1) }{ (n-1)! } +\cfrac { 3 }{ 6 } { \sum }_{ i=2 }^{ \infty }\cfrac { 1 }{ (n-1)! } \\ S=\cfrac { 2 }{ 6 } { \sum }_{ i=2 }^{ \infty }\cfrac { 1 }{ (n-2)! } +\cfrac { 1 }{ 2 } { \sum }_{ i=2 }^{ \infty }\cfrac { 1 }{ (n-1)! } \\ S=\cfrac { 1 }{ 3 } e+\cfrac { 1 }{ 2 } (e-1)\\ S=\cfrac { 5e }{ 6 } -\cfrac { 1 }{ 2 }$

Let

$S$ denote the sum of the infinite series

$1+\cfrac { 8 }{ 2! } +\cfrac { 21 }{ 3! } +\cfrac { 40 }{ 4! } +\cfrac { 65 }{ 5! } +.......$ . Then

0%
$S< 8$
0%
$S> 12$
0%
$8< S< 12$
0%
$S=8$

Explanation

$\; \;\; S_n=1+8+21+40+65+.....+a_n \\ - \underline{ S_n=\quad \; \;1+\;8+21+40+.....+a_{n-1}+a_n }$

$\;\;\;0\;=1+7+13+19+25+...,,,,,,,,.-a_n$

${ a }_{ n }=1+7+13+19+25+....\\ { a }_{ n }=\cfrac { n }{ 2 } [2a+(n-1)d]=\cfrac { n }{ 2 } [2+(n-1)6]\\ { a }_{ n }=n(3n-2)\\ S={ \sum }_{ i=1 }^{ \infty }\cfrac { n(3n-2) }{ n! } ={ \sum }_{ i=1 }^{ \infty }\cfrac { (3n-2) }{ (n-1)! }$

Adding and subtracting

$1$ in numerator

$S={ \sum }_{ i=2 }^{ \infty }\cfrac { 3(n-1) }{ (n-1)! } +{ \sum }_{ i=1 }^{ \infty }\cfrac { 1 }{ (n-1)! } \\ S=3{ \sum }_{ i=2 }^{ \infty }\cfrac { 1 }{ (n-2)! } +{ \sum }_{ i=1 }^{ \infty }\cfrac { 1 }{ (n-1)! } \\ S=3e+e\\ S=4e\\ 8<S<12\\ { e }^{ x }=1+\cfrac { x }{ 1! } +\cfrac { { x }^{ 2 } }{ 2! } +\cfrac { { x }^{ 3 } }{ 3! } +\cfrac { { x }^{ 4 } }{ 4! } +....\infty \\ e=1+\cfrac { 1 }{ 1! } +\cfrac { 1 }{ 2! } +\cfrac { 1 }{ 3! } +\cfrac { 1 }{ 4! } +....\infty$

$x=1+\frac{1}{2\times \underline{|1}}+\frac{1}{4\times \underline{|2}}+\frac{1}{8\times \underline{|3}}$

0%
${e}^{1/2}$
0%
$e^2$
0%
e
0%
$\frac{1}{e}$

Explanation

We know,

$e^{x}=1+\dfrac{x}{1!}+\dfrac{x^{2}}{2!}+\dfrac{x^{3}}{3!}+...$

put

$x=\dfrac{1}{2}$

$\therefore e^{\frac{1}{2}}=1+\dfrac{1}{2\times1!}+\dfrac{1}{4\times2!}+\dfrac{1}{8\times3!}+...$

$\therefore Ans=\sqrt{e}$

Find the

$(2n)^{th}$ term of the series whose

$n^{th}$ term is

$\dfrac{n^2+1}{n^3}$ :

0%
$\dfrac{n^2+1}{8n^3}$
0%
$\dfrac{4n^2+1}{8n^3}$
0%
$\dfrac{4n^2+1}{n^3}$
0%
$\dfrac{2n^2+1}{2n^3}$

Explanation

${ T }_{ 2n }=\cfrac { { (2n) }^{ 2 }+1 }{ { (2n) }^{ 3 } } =\cfrac { 4{ n }^{ 2 }+1 }{ 8{ n }^{ 3 } }$

In the series

$1+3+6+10+......,$ find the

$n^{th}$ term:

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

Explanation

$1=1$ , here $n=1\Rightarrow \dfrac {1(1+1)}{2}=1$
$1+2=3$ , here $n=2\Rightarrow \dfrac {2(2+1)}{2}=3$

$1+2+3=6$ , here $n=3\Rightarrow \dfrac {3(3+1)}{2}=6$

$1+2+3+4=10$ , here $n=4\Rightarrow \dfrac {4(4+1)}{2}=10$

${ n }^{ th }$ term is sum of $n$ terms

Therefore, ${ T }_{ n }=\dfrac { n(n+1) }{ 2 }$

Consider an incomplete pyramid of balls on a square base having

$18$ layers; and having

$13$ balls on each side of the top layer. Then the total number

$N$ of balls in that pyramid satisfies

0%
$9000 < N < 10000$
0%
$8000 < N < 9000$
0%
$7000 < N < 8000$
0%
$10000 < N < 12000$

Explanation

The top layer has $(13\times 13)$ balls the layer below it will have $(14\times 14)$ balls

We have $18$ layers

So the total number of balls

$N=(13\times 13)+(14\times 14)+.......(30\times 30)\\ N={ 13 }^{ 2 }+{ 14 }^{ 2 }+.....{ 30 }^{ 2 }$

Sum of squares of first $n$ natural numbers is $\dfrac { n(n+1)(2n+1) }{ 6 } \\$

$\therefore N=$ sum of first $30$ $-$ sum of first $12$

$N=\dfrac { 30\times 31\times 61 }{ 6 } -\dfrac { 12\times 13\times 25 }{ 6 } \\ N=8805\\$

$\Rightarrow 8000<N<9000$

Hence, option B is correct.

With the help of match-sticks, Zalak prepared a pattern as shown below. When

$97$ matchsticks are used, the serial number of the figure will be ...........

0%
Figure 32
0%
Figure 95
0%
Figure 49
0%
Figure 48

Explanation

Pattern associated with the number of matchsticks:
Figure 1: No. of sticks

$=3=3+0$
Figure 2: No. of sticks

$=5=3+2$
Figure 3: No. of sticks

$=7=3+2+2$
Figure 4: No. of sticks

$=9=3+2+2+2$
----------------------------------------
The rule associated with it can be given as no. of sticks

$=3+2(n-1)$ where

$n$ is the number of the figure.
Now there are

$97$ matchsticks

$\implies 3+2(n-1)=97$

$\implies 3+2n-2=97$

$\implies 2n=96$

$\implies n=48$

Hence, serial number of the figure having

$97$ matchsticks is

$48$ .

Three bells commenced to toll at the same time and tolled at intervals of

$20, 30, 40$ seconds respectively. If they toll together at

$6$ am, then which of the following is the time at which they can toll together

0%
$6:55$ am
0%
$6:56$ am
0%
$6:57$ am
0%
$6:59$ am

Let

$a_n=\dfrac{4n+\sqrt{4n^2-1}}{\sqrt{2n+1}+\sqrt{2n-1}}$ then

$\displaystyle \sum_{n=1}^{144}a_n$ equals

0%
$2456$
0%
$2645$
0%
$2466$
0%
$2546$

Explanation

Given

${ a }_{ n }=\cfrac { 4n+\sqrt { 4{ n }^{ 2 }-1 } }{ \sqrt { 2n+1 } +\sqrt { 2n-1 } }$

Multiplying Dividing by

$\sqrt { 2n+1 } -\sqrt { 2n-1 }$

${ a }_{ n }=\cfrac { \left( 4n+\sqrt { 4{ n }^{ 2 }-1 } \right) \left( \sqrt { 2n+1 } -\sqrt { 2n-1 } \right) }{ \left( 2n+1 \right) -\left( 2n-1 \right) }$

${ a }_{ n }=\cfrac { 4n\sqrt { 2n+1 } -4n\sqrt { 2n-1 } +(2n+1)\sqrt { 2n-1 } -(2n-1)\sqrt { 2n+1 } }{ 2 }$

$a_n=\cfrac{(2n+1)\sqrt { 2n+1 } -(2n-1)\sqrt { 2n-1 } }{ 2 }$

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

$a_1=\cfrac { { \left( 3 \right) }^{ { 3 }/{ 2 } }-{ \left( 1 \right) }^{ { 3 }/{ 2 } } }{ 2 }$

$a_2=\cfrac { { \left( 5 \right) }^{ { 3 }/{ 2 } }-{ \left( 3 \right) }^{ { 3 }/{ 2 } } }{ 2 }$

$a_3=\cfrac { { \left( 7\right) }^{ { 3 }/{ 2 } }-{ \left( 5 \right) }^{ { 3 }/{ 2 } } }{ 2 }$

$a_n=\cfrac { { \left( 2n+1 \right) }^{ { 3 }/{ 2 } }-{ \left( 2n-1 \right) }^{ { 3 }/{ 2 } } }{ 2 }$

$\sum _{ i=1 }^{n }{ { a }_{ n } }=\cfrac { { \left( 2n+1 \right) }^{ { 3 }/{ 2 } }-1 }{ 2 }$

$\sum _{ i=1 }^{144 }{ { a }_{ n } }$

$=\cfrac { { \left( 2\times 144+1 \right) }^{ { 3 }/{ 2 } }-1 }{ 2 }=\cfrac { { \left( 289 \right) }^{ { 3 }/{ 2 } }-1 }{ 2 } \\ =\cfrac { { \left( 17\right) }^{ { 3 } }-1 }{ 2 }=2456$

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

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

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

0%
$\dfrac{\pi^2}{12}$
0%
$\dfrac{\pi^2}{24}$
0%
$\dfrac{\pi^2}{8}$
0%
$\dfrac{\pi^2}{4}$

Explanation

Let

$S=\sum ^{ \infty }{ \dfrac { 1 }{ { n }^{ 2 } } } =\dfrac { { \pi }^{ 2 } }{ 6 } \Rightarrow \dfrac { S }{ { 2 }^{ 2 } } =\dfrac { 1 }{ { 2 }^{ 2 } } +\dfrac { 1 }{ 4^{ 2 } } +\dfrac { 1 }{ 6^{ 2 } } +...\infty$

$\therefore { S }^{ ' }=S-\dfrac { S }{ 4 } =\dfrac { 3S }{ 4 } =\dfrac { 1 }{ { 1 }^{ 2 } } +\left(\dfrac { 1 }{ 2^{ 2 } } -\dfrac { 1 }{ 2^{ 2 } } \right)+\dfrac { 1 }{ 3^{ 2 } } +\left(\dfrac { 1 }{ 4^{ 2 } } -\dfrac { 1 }{ 4^{ 2 } } \right)+...\infty$

ie,

$\therefore { S }^{ ' }=\dfrac { 1 }{ { 1 }^{ 2 } } +\dfrac { 1 }{ 3^{ 2 } } +\dfrac { 1 }{ { 5 }^{ 2 } } +...\infty =\dfrac { 3S }{ 4 } =\dfrac { 3{ \pi }^{ 2 } }{ 4\times 6 } =\dfrac { { \pi }^{ 2 } }{ 8 }$

So, Option C is the correct one.

$\displaystyle \sum_{k = 1}^{\infty} \dfrac {6^{k}}{(3^{2k + 1} + 2^{2k + 1}) - (3^{k}2^{k + 1} + 2^{k}3^{k + 1})}$ is equal to

0%
$3$
0%
$\dfrac {1}{3}$
0%
$\dfrac {4}{3}$
0%
$\dfrac {6}{5}$

Explanation

${ \sum }_{ k=1 }^{ \infty }\cfrac { { 6 }^{ k } }{ ({ 3 }^{ 2k+1 }+{ 2 }^{ 2k+1 })-({ 3 }^{ k }{ 2 }^{ k+1 }+{ 2 }^{ k }{ 3 }^{ k+1 }) } \\ ={ \sum }_{ k=1 }^{ \infty }\cfrac { { 6 }^{ k } }{ 3\cdot { 3 }^{ 2k }-3\cdot { 3 }^{ k }{ 2 }^{ k }+{ 2 }^{ 2 }{ 2 }^{ 2k }-2\cdot { 3 }^{ k }{ 2 }^{ k } } \\ ={ \sum }_{ k=1 }^{ \infty }\cfrac { { 6 }^{ k } }{ 3\cdot { 3 }^{ k }({ 3 }^{ k }-{ 2 }^{ k })-{ 2 }\cdot { 2 }^{ k }({ 3 }^{ k }-{ 2 }^{ k }) } \\ ={ \sum }_{ k=1 }^{ \infty }\cfrac { { 6 }^{ k } }{ ({ 3 }^{ k }-{ 2 }^{ k })({ 3 }^{ k+1 }-{ 2 }^{ k+1 }) } \\ \cfrac { { 6 }^{ k } }{ ({ 3 }^{ k }-{ 2 }^{ k })({ 3 }^{ k+1 }-{ 2 }^{ k+1 }) } =\cfrac { { 3 }^{ k }A }{ ({ 3 }^{ k }-{ 2 }^{ k }) } +\cfrac { { 3 }^{ k }B }{ ({ 3 }^{ k+1 }-{ 2 }^{ k+1 }) } \\ \cfrac { { 6 }^{ k } }{ ({ 3 }^{ k }-{ 2 }^{ k })({ 3 }^{ k+1 }-{ 2 }^{ k+1 }) } =\cfrac { (3A+B){ 3 }^{ 2k }B-(2A+B){ 6 }^{ k } }{ ({ 3 }^{ k }-{ 2 }^{ k })({ 3 }^{ k+1 }-{ 2 }^{ k+1 }) } \\ 3A+B=0\Rightarrow 3A=-B\quad \quad (1)\\ 2A+B=-1\\ 2A-3A=-1\quad from(1)\\ -A=-1\\ A=1\Rightarrow B=-3A=-3\\ { \sum }_{ k=1 }^{ \infty }\cfrac { { 6 }^{ k } }{ ({ 3 }^{ k }-{ 2 }^{ k })({ 3 }^{ k+1 }-{ 2 }^{ k+1 }) } =\sum \cfrac { { 3 }^{ k } }{ ({ 3 }^{ k }-{ 2 }^{ k }) } -\sum \cfrac { { 3 }^{ k+1 } }{ ({ 3 }^{ k+1 }-{ 2 }^{ k+1 }) } \\ { T }_{ 1 }=\cfrac { 3 }{ 3-2 } -\cfrac { { 3 }^{ 2 } }{ { 3 }^{ 2 }-{ 2 }^{ 2 } } \\ { T }_{ 2 }=\cfrac { { 3 }^{ 2 } }{ { 3 }^{ 2 }-{ 2 }^{ 2 } } -\cfrac { { 3 }^{ 3 } }{ { 3 }^{ 3 }-{ 2 }^{ 3 } } \\ { T }_{ n }=\cfrac { { 3 }^{ k } }{ ({ 3 }^{ k }-{ 2 }^{ k }) } -\cfrac { { 3 }^{ k+1 } }{ ({ 3 }^{ k+1 }-{ 2 }^{ k+1 }) } \\ { S }_{ n }=3-\cfrac { { 3 }^{ n+1 } }{ { 3 }^{ n+1 }-{ 2 }^{ n+1 } } \\ \underset { n\rightarrow \infty }{ \lim } { S }_{ n }=3$

Let

$S = \displaystyle \sum_{n = 1}^{99} = \dfrac {5^{100}}{5^{100} + 25^{n}}$ then find the value of

$[S]$ , where

$[.] = G.I.F.$

0%
$99$
0%
$100$
0%
$25$
0%
$49$

The first term of an AP is

$148$ and the common difference is

$-2$ . If the AM of first

$n$ terms of the AP is

$125$ , then the value of

$n$ is

0%
$18$
0%
$24$
0%
$30$
0%
$36$
0%
$48$

Explanation

Given,

$a=148, d=-2$
AM of

$n$ terms

$=125$

$\Rightarrow \dfrac { { a }_{ 1 }+{ a }_{ 2 }+\cdots +{ a }_{ n } }{ n } =125$

$\Rightarrow \dfrac { \dfrac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] }{ n } =125$

$\Rightarrow 2a+\left( n-1 \right) d=250$

$\Rightarrow 2\times 148+\left( n-1 \right) \left( -2 \right) =250$

$\Rightarrow 296-2n+2=250$

$\Rightarrow 298-2n=250$

$\Rightarrow n=24$

The sum of the series

$\displaystyle \sum_{n = 8}^{17} \dfrac{1}{(n + 2)(n + 3)}$ is equal to

0%
$\dfrac{1}{17}$
0%
$\dfrac{1}{18}$
0%
$\dfrac{1}{19}$
0%
$\dfrac{1}{20}$
0%
$\dfrac{1}{21}$

Explanation

$\sum_{n=8}^{17}\cfrac1{(n+2)(n+3)}$

$=\sum_{n=8}^{17}\cfrac1{(n+2)}-$

$\cfrac1{(n+3)}$

$=\sum_{n=8}^{17}\cfrac1{(n+2)}-$

$\sum_{n=8}^{17}\cfrac1{(n+3)}$

$=\left(\cfrac1{10}+\cfrac1{11}+.....+\cfrac1{19}\right)-$

$\left(\cfrac1{11}+\cfrac1{12}+.....+\cfrac1{20}\right)$

$=\cfrac1{10}-\cfrac1{20}$

$=\cfrac1{20}$

Hence, D is the correct option.

The value of

$\dfrac {1}{i} + \dfrac {1}{i^{2}} + \dfrac {1}{i^{3}} + .... + \dfrac {1}{i^{102}}$ is

0%
$-1 - i$
0%
$-1 + i$
0%
$1 - i$
0%
$1 + i$
0%
$1 - 2i$

Explanation

$\dfrac {1}{i} + \dfrac {1}{i^{2}} + \dfrac {1}{i^{3}} + .... + \dfrac {1}{i^{102}}$ (GP)
Therefore,

$S_{n} = \dfrac {\dfrac {1}{i}\left \{1 - \left (\dfrac {1}{i}\right )^{102}\right \}}{1 - \left (\dfrac {1}{i}\right )} = \dfrac {(1 + 1)}{i - 1} \left (\because i^{2} = 1\ and\ i^{2} = -1\right )$

$= \dfrac {2(i + 1)}{i^{2} - 1} = -(i + 1) = -1 - i$

The sum of the first

$n$ terms of the series

${ 1 }^{ 2 }+2\cdot { 2 }^{ 2 }+{ 3 }^{ 3 }+2\cdot { 4 }^{ 2 }+{ 5 }^{ 2 }+2\cdot { 6 }^{ 2 }+\cdots$ is

$\dfrac { n{ \left( n+1 \right) }^{ 2 } }{ 2 }$ when

$n$ is even, when

$n$ is odd the sum is

0%
$\dfrac { 3n\left( n+1 \right) }{ 2 }$
0%
$\dfrac { { n }^{ 2 }\left( n+1 \right) }{ 2 }$
0%
$\dfrac { n{ \left( n+1 \right) }^{ 2 } }{ 4 }$
0%
${ \left[ \dfrac { n\left( n+1 \right) }{ 2 } \right] }^{ 2 }$

Explanation

The sum of

$n$ terms of given series

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

$n$ is odd, i.e.,

$n=2m+1$
Then,

${ S }_{ 2m+1 }\equiv { S }_{ 2m }+\left( 2m+1 \right)$ th term

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

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

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

Find the sum of the series

$\displaystyle \frac{1}{2\cdot 3}+\frac {1}{4\cdot 5}+\frac {1}{6\cdot 7}+ ...$

0%
$log\, \frac {e}{2}$
0%
$log\, \frac {e}{4}$
0%
$log\, \frac {2}{3}$
0%
$log\, \frac {2}{4}$

Explanation

$\displaystyle \frac {1}{2\cdot 3}+ \frac {1}{4\cdot 5}+\frac {1}{6\cdot 7}$

$\displaystyle \frac {1}{2}- \frac {1}{3}+ \frac {1}{4}- \frac {1}{5}+\frac {1}{6}-\frac {1}{7} + ...$

$\displaystyle =1-1 + \frac {1}{2}-\frac {1}{3}+ \frac {1}{4}- \frac {1}{5}+...$

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

$= 1- log \, 2 = log\, e - log\, 2$

$=log \frac {e}{2}$

If the natural numbers are divided into groups of {1}, {2, 3}, {4, 5, 6}, {7, 8, 9, 10} ....Then the /sum of 50th group is

0%
65225
0%
56225
0%
62525
0%
53625

Explanation

Pattern:

2 3

4 5 6

7 8 9 10

..........

Series for starting number:

1, 2, 4, 7, ...

Let

$T_{n}=an^{2}+bn+c$

a+b+c=1 ......(1)

4a+2b+c=2 ......(2)

9a+3b+c=4 .......(3)

From eq (1), (2), (3), we get:

$a=\dfrac{1}{2},b=-\dfrac{1}{2} ,c=1$

$\therefore T_{n}=\dfrac{n^{2}}{2}-\dfrac{n}{2}+1$

$\therefore T_{50}=\dfrac{50^{2}}{2}-\dfrac{50}{2}+1=1226$

Now, elements in 50th group:

1226, 1227, 1228,.....(50 terms)

$\therefore S_{n}=\dfrac{50}{2}[2\times 1226+(50-1)\times 1]=62525$

The value of a for which side of nth square equals the diagonals of

$(n + 1)^{th}$ square is

0%
1/3
0%
1/4
0%
1/2
0%
$1\sqrt{2}$

Explanation

Given

$(Diagonal)_{n+1}=(Side)_n \\ \sqrt{2}a_1r^n=a_1r^{n-1} \\ \sqrt{2}r=1 \\ r=\cfrac{1}{\sqrt{2}}$

$\displaystyle \sum_{r = 1}^{n}t_{n} = \dfrac {n(n +1)(n + 2)(n + 3)}{8}$ , then

$\displaystyle \sum_{r = 1}^{n} \dfrac {1}{t_{1}}$ equals

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

Let

$r^{th}$ term of a series is given by,

$T_r = \dfrac {r}{1-3r^2 + r^4} .$

Then

$\underset {n \rightarrow \infty}{\lim} \sum_{r=1}^n T_r$ is

0%
$\cfrac { 3 }{ 2 }$
0%
$\cfrac { 1 }{ 2 }$
0%
$\cfrac { -1 }{ 2 }$
0%
$\cfrac { -3 }{ 2 }$

Explanation

${ T }_{ r }=\cfrac { r }{ { ({ r }^{ 2 }-1) }^{ 2 }-{ r }^{ 2 } } =\cfrac { r }{ ({ r }^{ 2 }-1+r)({ r }^{ 2 }-1-r) }$

${ T }_{ r }=\cfrac { 1 }{ 2 } \cfrac { ({ r }^{ 2 }+r-1)-({ r }^{ 2 }-r-1) }{ ({ r }^{ 2 }-1+r)({ r }^{ 2 }-1-r) }$

${ T }_{ r }=\cfrac { 1 }{ 2 } \left[ \cfrac { 1 }{ ({ r }^{ 2 }-1-r) } -\cfrac { 1 }{ ({ r }^{ 2 }-1+r) } \right]$

${ T }_{ 1 }=\cfrac { 1 }{ 2 } \left[ -1-\cfrac { 1 }{ 1 } \right]$

${ T }_{ 2 }=\cfrac { 1 }{ 2 } \left[ \cfrac { 1 }{ 1 } -\cfrac { 1 }{ 5 } \right]$

${ T }_{ 3 }=\cfrac { 1 }{ 2 } \left[ \cfrac { 1 }{ 5 } -\cfrac { 1 }{ 11 } \right]$

${ T }_{ n }=\cfrac { 1 }{ 2 } \left[ \cfrac { 1 }{ ({ n }^{ 2 }-1-n) } -\cfrac { 1 }{ ({ n }^{ 2 }-1+n) } \right]$

Adding all we get

${ S }_{ n }=\cfrac { 1 }{ 2 } \left[ -1-\cfrac { 1 }{ ({ n }^{ 2 }-1-n) } \right]$

$\underset { n\rightarrow \infty }{ \lim } { \sum }_{ r=1 }^{ n }{ T }_{ n }=\underset { n\rightarrow \infty }{ \lim } \cfrac { 1 }{ 2 } \left[ -1-\cfrac { 1 }{ ({ n }^{ 2 }-1-n) } \right] =\cfrac { -1 }{ 2 } (1-0)=\cfrac { -1 }{ 2 }$

$\alpha =1 / 4$ and

$P_n$ denotes the perimeter of the nth square then

$\sum_{n=1}^{\infty } P_n$ equals

0%
$\dfrac{8}{3}\left ( 4+\sqrt{10} \right )$
0%
$\dfrac{8}{3}$
0%
$\dfrac{16}{3}$
0%
None of these

Explanation

Parameter of

$n^{th}$ square

$=4a_n=4a_1r^{n-1}$

$\sum_{n=1}^{\infty}$

$=P_1+P_2+P_3+....+P_n \\ =4a_1(1+r+r^2+r^3+....infty)$

Sum of infinite GP

$=\cfrac{a}{1-r}$

$\sum_{n=1}^{\infty}=4a_1\cfrac{1}{1-r}=\cfrac{4a_1}{1-\sqrt{2\alpha ^2-2/alpha+1}}$

Given

$a_1=1, \quad \alpha=\cfrac{1}{4}$

$\sum_{n=1}^{\infty}$

$=\cfrac { 4 }{ 1-\sqrt { \cfrac { 2 }{ 16 } -\cfrac { 2 }{ 4 } +1 } } =\cfrac { 4 }{ 1-\cfrac { \sqrt { 2-8+16 } }{ 4 } } =\cfrac { 16 }{ 4-\sqrt { 10 } } \\ =\cfrac { 16 }{ 4-\sqrt { 10 } } \times \cfrac { 4+\sqrt { 10 } }{ 4+\sqrt { 10 } } =\cfrac { 16(4+\sqrt { 10 } ) }{ 6 } =\cfrac { 8 }{ 3 } (4+\sqrt { 10 } )$

$a_1 \in R - \left \{ 0 \right \}, i = 1, 2, 3, 4$ and

$x \in R$ and

$\left ( \sum_{i=1}^{3}a_i^2\right ) x^2-2x \left ( \sum_{i=1}^{3}a_i a_{i+1}\right )+\sum_{i=2}^{4}a_i^2\leq 0$ , then

$a_1, a_2, a_3, a_4$ are in

0%
A.P
0%
G.P
0%
H.P
0%
A.G.P

Explanation

$({ { a }_{ 1 } }^{ 2 }+{ { a }_{ 2 } }^{ 2 }+{ { a }_{ 3 } }^{ 2 }){ x }^{ 2 }-2x({ a }_{ 1 }{ a }_{ 2 }+{ a }_{ 2 }{ a }_{ 3 }+{ a }_{ 3 }{ a }_{ 4 })+({ { a }_{ 2 } }^{ 2 }+{ { a }_{ 3 } }^{ 2 }+{ { a }_{ 4 } }^{ 2 })\le 0\\ ({ x }^{ 2 }{ { a }_{ 1 } }^{ 2 }-2x{ a }_{ 1 }{ a }_{ 2 }+{ { a }_{ 2 } }^{ 2 })+({ { a }_{ 2 } }^{ 2 }{ x }^{ 2 }-2{ a }_{ 2 }{ a }_{ 3 }+{ { a }_{ 3 } }^{ 2 })+({ { a }_{ 3 } }^{ 2 }{ x }-2x{ a }_{ 3 }{ a }_{ 4 }+{ { a }_{ 4 } }^{ 2 })\le 0\\ { ({ a }_{ 1 }x-{ a }_{ 2 }) }^{ 2 }+{ ({ a }_{ 2 }x-{ a }_{ 3 }) }^{ 2 }+{ ({ a }_{ 3 }x-{ a }_{ 4 }) }^{ 2 }\le 0$

Since each one is a square term

These sum of term cannot be less than

$0$

Only possibility they can be zero

For

$a_1x-a_2=0 \Rightarrow x=\cfrac{a_2}{a_1} \\ a_2x-a_3=0 \Rightarrow x=\cfrac{a_3}{a_2} \\ a_3x-a_4=0 \Rightarrow x=\cfrac{a_4}{a_3} \\ x=\cfrac{a_2}{a_1} = \cfrac{a_3}{a_2} = \cfrac{a_4}{a_3}=k$

Hence,

$a_1,a_2,a_3,a_4$ are in GP

${ b }_{ i }=1-{ a }_{ i },na=\sum _{ i=1 }^{ n }{ { a }_{ i } } ,nb=\sum _{ i=1 }^{ n }{ { b }_{ i } } \quad$ , then

$\sum _{ i=1 }^{ n }{ { { a }_{ i }b }_{ i } } +\sum _{ i=1 }^{ n }{ { \left( { a }_{ i }-a \right) }^{ 2 } } =$

0%
$ab$
0%
$-nab$
0%
$nab$
0%
$(n+1)ab$

Explanation

$a_1+b_1=1 \quad \quad na=a_1+a_2+a_3+...+a_n \quad (1) \\ a_2+b_2=1 \quad \quad nb=b_1+b_2+b_3+....+b_n \quad (2) \\ a_3+b_3=1 \\ ----- \\ \underline{a_n+b_n=1} \\ (a_1+a_2+a_3+...+a_n)+(b_1+b_2+b_3+....+b_n)=n \quad (3)$

From

$(1),(2),(3)$

$na+nb=n \\ a+b=1 \quad \quad (4)$

${ \sum }_{ i=1 }^{ n }{ a }_{ i }{ b }_{ i }+{ \sum }_{ i=1 }^{ n }{ ({ a }_{ i }-a) }^{ 2 }\\ \Rightarrow { \sum }_{ i=1 }^{ n }{ a }_{ i }(1-{ a }_{ i })+{ \sum }_{ i=1 }^{ n }{ { a }_{ i } }^{ 2 }+{ \sum }_{ i=1 }^{ n }{ a }^{ 2 }-2{ \sum }_{ i=1 }^{ n }{ a }_{ i }a\\ \Rightarrow { \sum }_{ i=1 }^{ n }{ a }_{ i }-{ \sum }_{ i=1 }^{ n }{ { a }_{ i } }^{ 2 }+{ \sum }_{ i=1 }^{ n }{ { a }_{ i } }^{ 2 }+{ a }^{ 2 }{ \sum }_{ i=1 }^{ n }(1)-2a{ \sum }_{ i=1 }^{ n }{ a }_{ i }\\ =(1-2a){ \sum }_{ i=1 }^{ n }{ a }_{ i }+{ a }^{ 2 }{ \sum }_{ i=1 }^{ n }(1)\\ =(1-2a)(na)+{ a }^{ 2 }(n)\\ =na-2{ a }^{ 2 }n+{ a }^{ 2 }n\\ =na-{ a }^{ 2 }n\\ =na(1-a)\\ =nab\quad \quad from(4)$

$4,9,25,?,121,169$

0%
$36$
0%
$49$
0%
$64$
0%
$81$

Explanation

Given series is

$4,9,25,?,121, 168$

It can be written as

${2}^{2},{3}^{2},{5}^{2},{7}^{2},{11}^{2},{13}^{2}$

So, the missing term would be

$7^2$ i.e.

$49$ .

Let S be the infinite sum given by

$S=\displaystyle \sum_{n=0}^{\infty}\frac{a_n}{10^{2n}}$ , where

$(a_n)_{n\geq 0}$ is a sequence defined by

$a_0=a_1=1$ and

$a_j=20a_{j-1}$ for

$j\geq 2$ . If

$S$ is expressed in the form

$\displaystyle\frac{a}{b}$ , where

$a, b$ are coprime positive integers, than

$a$ equals.

0%
$60$
0%
$75$
0%
$80$
0%
$81$

Explanation

Given,

$S = \sum_{n=0}^\infty \dfrac{a_n}{10^{2n}}$

Expanding the summation, we get

$S = \dfrac{a_0}{1}+\dfrac{a_1}{10^2}+\dfrac{a_2}{10^4}+\dfrac{a_3}{10^6}+\dfrac{a_4}{10^8}+\dots$

On substituting the given values, we have

$S = 1+\dfrac{1}{10^2}+\dfrac{20}{10^4}+\dfrac{20^2}{10^6}+\dfrac{20^3}{10^8}+\dots$

$S = 1+\dfrac{1}{10^2}+\dfrac{2}{10^3}+\dfrac{2^2}{10^4}+\dfrac{2^3}{10^5}+\dots$

$S = 1+\dfrac{1}{10^2}\left\lbrace 1 + \dfrac{2}{10}+\dfrac{2^2}{10^2}+\dfrac{2^3}{10^3}+\dots\right\rbrace$

$S = 1+\dfrac{1}{10^2}\left\lbrace 1 + \dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+\dots\right\rbrace$

Since the terms in the bracket form an infinite geometric progression, we can sum them as

$S = 1+\dfrac{1}{10^2}\left\lbrace \dfrac{1}{1-\dfrac{1}{5}} \right \rbrace$

$S = 1+\dfrac{1}{10^2}\left\lbrace \dfrac{5}{4} \right \rbrace$

$S = 1+\dfrac{5}{400}$

$S = \dfrac{405}{400}$

As we have coprime numerator and denominator requirement,

$\dfrac{a}{b} = \dfrac{81}{80}$

Hence, the required answer is

$81$ .

Find the missing number in the circle:

0%
$66$
0%
$72$
0%
$71$
0%
$78$

Explanation

$3\times 2 - 1, 5\times 2 - 1, 9\times 2 - 1, 17\times 2 - 1 + 39 = 72$ .

$\therefore$ The solution is

$72$ .

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

0%
$\displaystyle \frac{3^{n+2} - 2n + 5}{(n+1)(n+2)}$
0%
$\displaystyle \frac{3^{n+2} - 4n + 5}{(n+1)(n+2)}$
0%
$\displaystyle \frac{3^{n+2} - 2n - 5}{(n+1)(n+2)}$
0%
None of these

13, 74, 290, 650,.......

0%
$1248$
0%
$1370$
0%
$1346$
0%
$1452$
0%
$1625$

Explanation

Here we observe the first term 13 can be written as:

$2^2+3^2$

Similarly other numbers can be written as:

$74: 5^2+7^2$

$290: 11^2+13^2$

$650: 17^2+19^2$

Here, we observe that all above term of series can be written as sum of squares of two prime numbers which comes next to it's previous pair.

Like:

$(2,3);(5,7);(11,13);(17,19)$ So, other pair of prime number will be

$(23,29).$

Hence next term of given series will be:

$\Rightarrow 23^2+29^2=1370$

So, correct answer is

$1370$ .

Figures

$1$ and

$2$ are related in a particular manner. Establish the same relationship between figures

$3$ and

$4$ by choosing a figure from amongst the options.

Select the INCORRECT match

0%
$3249-MMMCCXLIX$
0%
$1667-MDCLXVII$
0%
$207-CCXVII$
0%
$499-CDXCIX$

Explanation

(a)

$MMMCCXLIX=3000+200+40+9=3249$
(b)

$MDCLXVII=1000+500+100+60+7=1667$
(c)

$CCXVII=217=200+17$
(d)

$CDXCIX=400+90+9=499$

Hence the correct match is option C.

The value of

$\sum _{ n=1 }^{ \infty }{ { \left( -1 \right) }^{ n+1 }\left( \cfrac { n }{ { 5 }^{ n } } \right) }$ equals

0%
$\cfrac { 5 }{ 12 }$
0%
$\cfrac { 5 }{ 24 }$
0%
$\cfrac { 5 }{ 36 }$
0%
$\cfrac { 5 }{ 16 }$

If the

$p^{th}$ term of the series of positive numbers

$25, 22\dfrac {3}{5}, 20\dfrac {1}{2}, 18\dfrac {1}{4}$ , .... is numerically the smallest, then the

$p^{th}$ is.

0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{7}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{5}$

Explanation

$d = 22\dfrac{3}{4} - 25$

$=-2\dfrac{1}{4} = -\dfrac{9}{4}$

$t_n \ge 0$

$\therefore t_n = 25 + (n-n)\left(-\dfrac{9}{4}\right)$

$25-\dfrac{9}{4} (n-1) \ge 0$

$\Rightarrow 25\ge \dfrac{9}{4} (p-1) \ge 0$

$\Rightarrow \dfrac{100}{9} \ge p -1$

$\Rightarrow \dfrac{109}{9} \ge p$

$12 \dfrac{1}{9} \ge p$
12th term is positive.

$25 + (12 - 1) \times \left(-\dfrac{9}{4}\right) = \dfrac{1}{4}$

The sum of first

$20$ terms of the series

$1,6,13,22$ - is

0%
$5580$
0%
$5780$
0%
$7789$
0%
$1237$

Explanation

Given series,

$S = 1,6,13,22,...$

Sum of series,

$S_{n} = 1+6+13+22+...+ T_{n}$

$S_{n} = 1+6+13+...+T_{n}+T_n$

$0 = 1+5+7+9+,,,T_{n}$

$T_{n} = 1+ [5+7+9+...(n-1)terms]$

(sum of AP)

$= 1+\left [ \left ( \dfrac{n-1}{2} \right )(2(5) + (n-2)2) \right ]$

$= 1+ \left [ (n-1)(n-3) \right ]$

$= (n+1)^2 -3$

sum of 20 terms in series

$= \displaystyle \sum_{n=1}^{20} (n^2 + 2n-2)$

$= \displaystyle \sum_{n=1}^{20}n^2 +2\sum_{n=1}^{20} n - 2 \times 20$

$= 5780$

1161185_790379_ans_b3577ac2536e40bc8d1d192c9219af06.jpg

$\begin{vmatrix} x \end{vmatrix}<1$ then the coefficient of

$x^5$ in the expansion of

$\dfrac{3x}{(x-2) (x-1)}$ is

0%
$\dfrac{33}{32}$
0%
$-\dfrac{33}{32}$
0%
$\dfrac{31}{32}$
0%
$-\dfrac{33}{34}$

Explanation

$\dfrac{3x}{(x-2) (x+1)}= \dfrac {A}{x-2}+\dfrac{B}{x+1}$

$A=\dfrac{3(2)}{2+1}=2; B=\dfrac{3(-1)}{-1-2}=\dfrac{-3}{-3}=1$

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

$=\dfrac{2}{2(1-\dfrac {x }{ 2 } )} +\dfrac {1}{x+1} = -\begin{pmatrix} 1-\dfrac { x }{ 2 } \end{pmatrix}^{-1} +(1+x)^{-1}$

$=-\begin{pmatrix} 1+\dfrac{x}{2}+\begin{pmatrix} \dfrac {x}{2}\end{pmatrix}^2+......+\begin{pmatrix} \dfrac{x}{2} \end{pmatrix}^5+...... \end{pmatrix}+(1-x+x^2-x^3+x^4-x^5+......)$

Coefficient of

$x^5=\dfrac{-1}{32}-1=\dfrac{-33}{32}$

$2.4+4.7+6.10+.....$ upto

$(n-1)$ terms

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

Explanation

The general term of the series is

$T_n=2n(3n+1)=6n^2+2n$

The sum of series to

$n$ terms is

$6\times\dfrac{n(n+1)(2n+1)}6+2\times\dfrac{n(n+1)}2=n(n+1)[(2n+1)+1]=2n(n+1)^2$

Now, we need the sum till

$n-1$ terms, so we substitute

$n-1$ in place of

$n$

Hence, sum of series is

$2(n-1)n^2=2n^3-2n^2$

Sum of the series

$\sum _{ r=1 }^{ n }{ \left( { r }^{ 2 }+1 \right) r! }$ is ______

0%
$(n+1)!$
0%
$(n+2)!-1$
0%
$n(n+1)!$
0%
none of these

Explanation

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

$\Rightarrow \sum_{ r=1 }^{ n }{ ( { r }^{ 2 }+1+2r-2r ) r! }$

$\Rightarrow \sum_{ r=1 }^{ n }{ ( { ( r+1) }^{ 2 }.r!-2r.r! ) }$

$\Rightarrow \sum_{ r=1 }^{ n }{ ( { ( r+1) }^{ }.(r+1)!-2r.r! ) }$

$\Rightarrow \sum_{ r=1 }^{ n }{ [ ( r+2-1 )( r+1 )!-2( ( r+1-1 )r! ) ] }$

$\Rightarrow \sum_{ r=1 }^{ n }{ [ ( r+2 )( r+1 )!-( r+1 )!-2( ( r+1 )r!-r! ) ] }$

$\Rightarrow \sum_{ r=1 }^{ n }{ \left [ ( r+2 )!-( r+1 )!-2.( r+1 )!+2.r!) \right] }$

$T_1=3!-2!- 2.2!+2.1!$

$T_2=4!-3!-2.3!+2.2!$

$T_3=5!-4!-2.4!+2.3!$

$T_n=(n+2)!-(n+1)!-2.(n+1)!+2.n!$

$-------------------$

$Sum=S_n=(n+2)!-2!-2(n+1)!+2.1!$

$S_n=(n+2)!-2(n+1)!-2+2$

$S_n=(n+1)![n+2-2]$

$S_n=n.(n+1)!$

Correct answer is C

Given

$S_n=1+q+q^2+...+q^n$ &

$S_n=1+\displaystyle\frac{q+1}{2}+\left(\frac{q+1}{2}\right)^2+....+\left(\displaystyle\frac{q+1}{2}\right)^n, q\neq 1$ . then

$^{n+1}C_1+ {^{n+1}C_2\cdot s_1}+ {^{n+1}C_3}\cdot s_2+....+{^{n+1}C_{n+1}}\cdot s_n=2^n\cdot S_n$ .

0%
True
0%
False

Suppose

$a_2,a_3,a_4,a_5,a_6,a_7$ are integers such that

$\dfrac {5}{7}=\dfrac {a_2}{2!}+\dfrac {a_3}{3!}+\dfrac {a_4}{4}+\dfrac {a_5}{5!}+\dfrac {a_6}{6!}+\dfrac{a_7}{7!}$
where

$0 \le a < j$ for

$j=2,4,5,6,7.$ The sum

$a_2+a_3+a_4+a_5+a_6+a_7$ is

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

Explanation

$\cfrac { 5 }{ 7 } =\cfrac { { a }_{ 2 } }{ 2! } +\cfrac { { a }_{ 3 } }{ 3! } +\cfrac { { a }_{ 4 } }{ 4! } +\cfrac { { a }_{ 5 } }{ 5! } +\cfrac { { a }_{ 6 } }{ 6! } +\cfrac { { a }_{ 7 } }{ 7! } \\ \Rightarrow \cfrac { 5 }{ 7 } =\cfrac { 1 }{ 2 } \left( { a }_{ 2 }+\cfrac { 1 }{ 3 } \left( { a }_{ 3 }+\cfrac { 1 }{ 4 } \left( { a }_{ 4 }+......\left( \cfrac { { a }_{ 7 } }{ 7 } \right) \right) \right) \right) \\ \cfrac { 10 }{ 7 } ={ a }_{ 2 }+\cfrac { 1 }{ 3 } \left( { a }_{ 3 }+\cfrac { 1 }{ 4 } \left( { a }_{ 4 }+......\left( \cfrac { { a }_{ 7 } }{ 7 } \right) \right) \right) \\ 1+\cfrac { 3 }{ 7 } ={ a }_{ 2 }+\cfrac { 1 }{ 3 } \left( { a }_{ 3 }+\cfrac { 1 }{ 4 } \left( { a }_{ 4 }+......\left( \cfrac { { a }_{ 7 } }{ 7 } \right) \right) \right)$

So, as expression is less than $1$

${ a }_{ 2 }=1\\ \cfrac { 3 }{ 7 } =\cfrac { 1 }{ 3 } \left( { a }_{ 3 }+\cfrac { 1 }{ 4 } \left( { a }_{ 4 }+......\cfrac { 1 }{ 5 } \left( { a }_{ 5 }+.....\cfrac { { a }_{ 7 } }{ 7 } \right) \right) \right) \\ 1+\cfrac { 2 }{ 7 } ={ a }_{ 3 }+\cfrac { 1 }{ 4 } \left( { a }_{ 4 }+......\cfrac { 1 }{ 5 } \left( { a }_{ 5 }+\cfrac { 1 }{ 6 } \left( { a }_{ 6 }+\cfrac { { a }_{ 7 } }{ 7 } \right) \right) \right) \\ { a }_{ 3 }=1$

Similarly

${ a }_{ 4 }=1\quad \quad { a }_{ 7 }=2\\ { a }_{ 5 }=0\\ { a }_{ 6 }=4$

So, Sum ${ a }_{ 7 }-1+1+1+4+2\\ =9$

CBSE Questions for Class 11 Engineering Maths Sequences And Series Quiz 6 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Sequences And Series Quiz 6 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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