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

If $$\displaystyle \frac{a_{1}^{2}+a_{2}^{2}+...+a_{11}^{2}}{11}=90 $$, 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

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

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

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

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

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

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

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

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

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

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 $$n\rightarrow \infty $$

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

Explanation

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

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

if $$\displaystyle \frac{1}{1^2}\, +\, \displaystyle \frac{1}{2^2}\, +\, \displaystyle \frac{1}{3^2}$$ + .......... upto $$\infty\, =\, \displaystyle \frac{\pi^2}{6}$$, then $$\displaystyle \frac{1}{1^2}\, +\, \displaystyle \frac{1}{3^2}\, +\, \displaystyle \frac{1}{5^2}$$ + ........... = .......... .

0%
$$\pi^2 / 8$$
0%
$$\pi^2 / 12$$
0%
$$\pi^2 / 3$$
0%
$$\pi^2 / 9$$

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

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

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

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

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

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. $$a_1, a_2 ...... a_{2013}$$ 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?

0%
$$(-1)^x = 1; (-1)^y = 1$$
0%
$$(-1)^x = 1; (-1)^y = -1$$
0%
$$(-1)^x = -1; (-1)^y = 1$$
0%
$$(-1)^x = -1; (-1)^y = -1$$

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

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

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

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

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%
$$\sin { \left( \cfrac { \pi }{ 180 } \right) } +\sin { \left( \cfrac { \pi }{ 360 } \right) } $$

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%
$$\dfrac{1}{999}$$

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

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%
$$\dfrac{13}{12}$$

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

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%
$$2014-\cfrac { 1 }{ 2014 } $$

If $$ s_n = \displaystyle \sum_{r=0}^n \dfrac {1}{^nC_r}$$ 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%
$$\dfrac{2n - 1}{2}$$

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%
$$\dfrac {3n+5}{3n+1}$$

If $$\displaystyle \sum _{ r =1 }^{n}{ t_r = \frac{n(n+1)(n+2)(n+3)}{8} }$$, 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%
$$\displaystyle \left( \frac{1}{(n-1)(n-2)} + \frac{1}{2} \right) $$

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%
$$9999$$

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

If $$x\in R$$ 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%
$$\cfrac { 1948 }{ \sqrt { { 2 }^{ 1947 } } } $$

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

If $$x=\dfrac{1}{5}+\dfrac{1.3}{5.10}+\dfrac{1.3.5}{5.10.15}+.....\infty$$ 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%
$$\dfrac { 2{ \left( m+1 \right) }^{ 2 } }{ 1+{ \left( m+1 \right) }^{ 2 } }$$

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

If $$a_{1}, a_{2}, ......., a_n(n > 3)$$ 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%
$$E > 2^{n}$$

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

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%
$$513$$

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

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

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

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

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

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

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

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

If $$\dfrac{x}{0.2} + \dfrac{x}{0.3} + \dfrac{x}{0.6} + \dfrac{x}{0.4} + \dfrac{x}{0.5} = 87$$, 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

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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