MCQ Questions for Class 11 Engineering Maths Sequences And Series Quiz 8

The value of

$S=\sum _{ k=1 }^{ 6 }{ \left( \sin { \dfrac { 2\pi k }{ 7 } } -i\cos { \dfrac { 2\pi k }{ 7 } } \right) }$ ?

0%
$-1$
0%
$0$
0%
$-i$
0%
$i$

Explanation

$sin \dfrac{2 \pi k}{7} - i cos\dfrac{2 \pi k}{7}$

$= -i \left ( cos\dfrac{2\pi k}{7}+i sin \dfrac{2 \pi k}{7} \right )$

$= -ie^{2\pi k/7i}$

$\displaystyle \Rightarrow S = \sum_{k=1}^{6} -i\left ( e^{2\pi k/i} \right )$

$\displaystyle = -i \left ( \sum_{k=1}^{6} e^{i2\pi k/7} \right )$

$= -i$ [sum of roots of the equation

$x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+1 = 0].$

$= -i \left ( -1 \right )$

$= i$

1177083_1039505_ans_0d7a47a0f59e4258b0c3e756c7ebe7e8.jpg

The value of
$\frac{1}{1!} + \frac{1+2}{2!} + \frac{1+2+3}{3!} + \frac{1+2+3+4}{4!}+ \cods$ upto infinite terms is equal to

0%
(a) $\frac{3e}{2}$
0%
(b) $3e -1$
0%
(c) $2e +1$
0%
(d) $\frac{5e}{2}$

Find sum

${1}^{2}+{3}^{2}+{5}^{2}+...+{(2n-1)}^{2}$

0%
$\dfrac{n(2n+1)(2n+1)}{3}$ .
0%
$\dfrac{n(2n-1)(2n+1)}{3}$ .
0%
$\dfrac{2n(2n-1)(2n+1)}{3}$ .
0%
None of these

Explanation

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

Here

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

Sum of the given series

$=$

$\sum\limits_{n = 1}^n {{a_n}}$

$= \sum\limits_{n = 1}^n {{{\left( {2n - 1} \right)}^2}}$

$= \sum\limits_{n = 1}^n {\left( {4{n^2} + 1 - 4n} \right)}$

$= 4\sum\limits_{n = 1}^n {{n^2}} + \sum\limits_{n = 1}^n 1 - 4\sum\limits_{n = 1}^n n$

$= \dfrac{{4n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + n - \dfrac{{4n\left( {n + 1} \right)}}{2}$

$= \dfrac{n}{6}\left[ {8{n^2} + 12n + 4 + 6 - 12n - 12} \right]$

$= \dfrac{n}{6}\left[ {8{n^2} - 2} \right]$

$= \dfrac{n}{3}\left[ {4{n^2} - 1} \right]$

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

The sum to the first n terms of the series

$\dfrac{1}{2} + \dfrac{3}{4} + \dfrac{7}{8} + \dfrac{{15}}{{16}} + {\rm{ }} \ldots$ is

$9 + {2^{ - 10}}$ . The value of n is

0%
$10$
0%
$9$
0%
$8$
0%
none of these

Explanation

$\dfrac{1}{2}+\dfrac{3}{4}+\dfrac{7}{8}+\dfrac{15}{16}+........$

$=\left ( 1-\dfrac{1}{2} \right )+\left ( 1-\dfrac{1}{4} \right )+\left ( 1-\dfrac{1}{8} \right)+....$

$=(1+1+1+.......upto\; n\; terms)-\left ( \dfrac{1}{2} +\dfrac{1}{4}+\dfrac{1}{8}+.....n\; terms\right )$

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

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

Now,

$n-1+2^{-n}=9+2^{-10}$

$\Rightarrow n=10$

$\sum\limits_{i = 1}^{20} {{{\sin }^{ - 1}}{\text{ }}{x_i} = 10\pi } then\;\sum\limits_{i = 1}^{20} {{x_i}} \;$ is equal to

0%
20
0%
10
0%
0
0%
1

What is the sum of the series

$\dfrac{1}{{{3^2} - 4}} + \dfrac{1}{{{7^2} - 4}} + \dfrac{1}{{{{11}^2} - 4}} + ..... + \dfrac{1}{{{{39}^2} - 4}}$

0%
$\dfrac{{10}}{{41}}$
0%
$\dfrac{{40}}{{41}}$
0%
$\dfrac{{20}}{{41}}$
0%
$\dfrac{{30}}{{41}}$

Explanation

$\begin{array}{l}\dfrac{1}{{{3^2} - 4}} + \dfrac{1}{{{7^2} - 4}} + \dfrac{1}{{{{11}^2} - 4}} + ..... + \dfrac{1}{{{{39}^2} - 4}}\\ = \dfrac{1}{{\left( {3 - 2} \right)\left( {3 + 2} \right)}} + \dfrac{1}{{\left( {7 - 2} \right)\left( {7 + 2} \right)}} + \dfrac{1}{{\left( {11 - 2} \right)\left( {11 + 2} \right)}} + ..... + \dfrac{1}{{\left( {39 - 4} \right)\left( {39 + 4} \right)}}\\ = \dfrac{1}{4}\left[ {\dfrac{{5 - 1}}{{5 \times 1}} + \dfrac{{9 - 5}}{{9 \times 5}} + \dfrac{{13 - 9}}{{13 \times 9}} + .....} \right]\\ = \dfrac{1}{4}\left[ {\dfrac{{5 - 1}}{{5 \times 1}} + \dfrac{{9 - 5}}{{9 \times 5}} + \dfrac{{13 - 9}}{{13 \times 9}} + .....} \right]\\ = \dfrac{1}{4}\left[ {1 - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{9} + \dfrac{1}{9} - \dfrac{1}{{13}} + .........\dfrac{1}{{37}} - \dfrac{1}{{41}}} \right]\\ = \dfrac{1}{4}\left[ {1 - \dfrac{1}{{41}}} \right] = \dfrac{{10}}{{41}}\end{array}$

Number of identical terms in the sequence

$2, 5, 8, 11,$ ___ upto

$100$ terms and

$3, 5, 7, 9, 11$ ____ upto

$100$ terms are

0%
$17$
0%
$33$
0%
$50$
0%
$147$

Explanation

Fierst series:

$2, 5, 8, 11.........$

$a_1 =2;\ d=3;\ n=100$

$l=20(100-1)^*5=2+99^*3=2+297=299$

so, series

$=2, 5, 8, 11,.....,197, 197,200,203,....,299$

second series;

$3, 5, 9, 11,.....$

$a_1=3;\ d=2;\ n=100$

$l=3+(100-1)^*2=3+99^*2=3+198=201$

so series

$=3, 5, 7, 9, 11, ...., 201$

series having similar terms;

$5, 11, 17, ....., 197$

$a_1=5;\ d=6;\ l=197$

$l=a_1+(n-1)d$

$197=5+(n-1)6$

$192/6=n-1$

$32+1=n$

$n=33$

The value of the expression

$\left(\dfrac{1}{2^2 -1}\right) + \left(\dfrac{1}{4^2 -1}\right) + \left(\dfrac{1}{6^2 -1}\right) + ... + \left(\dfrac{1}{20^2 -1}\right)$ is

0%
$\dfrac{9}{19}$
0%
$\dfrac{10}{19}$
0%
$\dfrac{10}{21}$
0%
$\dfrac{11}{21}$

Explanation

$n$ th term in the given series can be written as

$\dfrac{1}{(2n)^{2}-1}$

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

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

So, the given series can be written as

$\dfrac{1}{2}\left ( 1-\dfrac{1}{3} \right )+\dfrac{1}{2}\left ( \dfrac{1}{3} -\dfrac{1}{5}\right )+\dfrac{1}{2}\left ( \dfrac{1}{5} -\dfrac{1}{7}\right )+..............\dfrac{1}{2}\left ( \dfrac{1}{19} -\dfrac{1}{21}\right )$

$=\dfrac{1}{2}\left ( 1-\dfrac{1}{21} \right )$

$=\dfrac{10}{21}$

$67,84,95,.,133,158$

0%
$116$
0%
$129$
0%
$108$
0%
$111$

Let

$S_n=\dfrac{1}{1+2008n}+\dfrac{1}{2+2008n}+...+\dfrac{1}{2009n}$ . Then

$\displaystyle\lim_{n\rightarrow\infty}S_n$ equals?

0%
$log\left(\dfrac{1}{2008}\right)$
0%
$log\left(1+\dfrac{1}{2008}\right)$
0%
$log\left(\dfrac{1}{2009}\right)$
0%
$log\left(1+\dfrac{1}{2009}\right)$

Find the value of ?.

0%
$125$
0%
$25$
0%
$625$
0%
$156$

Explanation

${\left( {3 + 1} \right)^2} = {\left( 4 \right)^2} = 16$

${\left( {15 + 6} \right)^2} = {\left( {21} \right)^2} = 441$

${\left( {10 + 5} \right)^2} = {\left( {15} \right)^2} = 225$

$so,\,{\left( {12 + 13} \right)^2} = {\left( {25} \right)^2} = 625$

therefore the answer is

$625$

$\frac{a_2a_3}{a_1a_4}=\frac{a_2+a_3}{a_1+a_4}=3\left(\frac{a_2-a_3}{a_1-a_4}\right)$ then

$a_1, a_2$ and

$a_3$ are in which progression?

0%
A.P.
0%
G.P.
0%
H.P.
0%
None of these

$\sum n =55 \implies n=5$

0%
True
0%
False

Explanation

$\sum n=55\\\dfrac{n(n+1)}2=55\\n^2+n=110\\n^2+11n-10n-110=0\\n(n+11)-10(n+11)=0\\(n-10)(n+11)=0\\n=10$

The sum of all

$2$ -digit numbers divisible by

$5$ is _________?

0%
$1035$
0%
$1245$
0%
$1230$
0%
$945$

Explanation

The series will be : 10, 15, 20,.....,90,95

$S_{n} = 10 + 15 + 20 +...... + 90 + 95$

here a will be first term

a = 10

& d = difference b/w any

two terms

d = 5

series containing n terms

& n =

$1 + \dfrac{\left ( Last\, term - first \,term \right ) } {d}$

$= 1 + \dfrac{\left ( 95 - 10 \right )}{5}$

n = 1 + 17 = 18

now,

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

$= \dfrac{18}{2}\left [ 2\left ( 10 \right )+\left ( 18 - 1 \right )\left ( 5 \right ) \right ]$

$= 9\left ( 20 + 85 \right )$

$S_{n} = 9\left ( 105 \right )$

$\boxed { Sn = 945}$

1180629_1068790_ans_e6604095d8df4495a42b2eb38eba4e1b.jpg

$S_r=\left | \begin{array}{111} 2r & x & n(n+1)\\ 6r^2-1 & y & n^2(2n+3)\\ 4r^3-2nr & z & n^3(n+1)\\ \end{array} \right |$ then

$\sum_\limits{r=1}^n S_r$
does not depends on

0%
x
0%
y
0%
n
0%
all of these

$(1^2-t_1)+ (2^2-t_2) +.......+(n^2-1)$ , then

$t_n$ is

0%
$\frac{n}{2}$
0%
$n -1$
0%
$n+1$
0%
$n$

Insert the missing number in the given series :

$0,4,18,48, ?, 180$

0%
$58$
0%
$68$
0%
$84$
0%
$100$

Explanation

$0=0\times1^2$

$4=1\times2^2$

$18=2\times3^2$

$48=3\times4^2$

$\therefore4\times5^2=4\times25=100$ is the next term.

Ten students of the physics department decided to go on a educational trip.They hired a mini bus for the trip, but the bus can only carry eight students at a time and each student goes at least once. Find the minimum number of trips the bus has to make so that each students can go for equal number of trips.

0%
$5$
0%
$4$
0%
$6$
0%
$8$

Explanation

Ten students

$\longrightarrow$ Eight students in bus

$2$ students are excluded/not taken on a trip so we have exclude every students once, so that no. of trips made by each students are equal so no. of trips

$= \dfrac{10}{2} =5$

$A$ is correct.

For a sequence

$\left\{ { a }_{ n } \right\} ,{ a }_{ 1 }=2$ and

$\dfrac { { a }_{ n+1 } }{ { a }_{ n } } =\dfrac { 1 }{ 3 }$ . Then

$\sum _{ r=1 }^{ 20 }{ { a }_{ r } }$ is

0%
$\dfrac {20}{2}[4+19\times 3]$
0%
$3(1-\dfrac {1}{3^{20}})$
0%
$2(1-3^{20})$
0%
$(1-\dfrac {1}{3^{20}})$

Explanation

Given that,

$a_1=2$ , $r=\dfrac{a_{n+1}}{a_n}=\dfrac{1}{3}$

$\sum\nolimits_{r=1}^{20}{{{a}_{r}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}}$

$=\dfrac{2\left[ 1-{{\left( \dfrac{1}{3} \right)}^{20}} \right]}{1-\dfrac{1}{3}}$ $[n=20]$

$=\dfrac{2\left[ 1-{{\left( \dfrac{1}{3} \right)}^{20}} \right]}{\dfrac{2}{3}}$

$=3\left[ 1-{{\left( \dfrac{1}{3} \right)}^{20}} \right]$

Hence this is the correct answer.

Number of rectangles in the grid shown which are not squares is?

0%
$160$
0%
$162$
0%
$170$
0%
$185$

$1,2,1,4,3,8,9,5,27,16,?,?,?$

0%
$7,36,64$
0%
$7,25,64$
0%
$9,25,64$
0%
$7,25,49$

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

0%
a + c = b + d
0%
e = 0
0%
$a, b - \dfrac{2}{3}, \, c - 1$ are in A. P.
0%
$\dfrac{c}{a}$ is an integer

$\displaystyle \sum^{n}_{r=0} (-1)^r \,{^nC_r}. \dfrac{(1 + r \ell n 10)}{(1 + \ell n 10^n)^r} =$

0%
$0$
0%
$\dfrac{1}{2}$
0%
$1$
0%
$none\ of\ these$

Evaluate the definite integral:

$\displaystyle\int_{0}^{\pi/2}\dfrac{\cos^{2}x}{1+3\sin^{2}x}\ dx$

0%
$\pi /4$
0%
$\pi /2$
0%
$\pi /6$
0%
$\pi /3$

Explanation

Let

$I=\displaystyle\int_{0}^{\pi/2}\dfrac{\cos^{2}x}{1+3\sin^{2}x}\ dx$ . Then

$I=\displaystyle\int_{0}^{\pi/2}\dfrac{\cos^{2}x}{1+3\sin^{2}x}dx$ . Then,

$I=\displaystyle\int_{0}^{\pi/2}\dfrac{\sec^{2}x}{\sec^{2}x(\sec^{2}x+3\tan^{2}x)}\ dx$ [Diving

$Numerator$ and

$Denominator$ by

$\cos^{4}x$ ]

$\Rightarrow I=\displaystyle\int_{0}^{\infty}\dfrac{1}{(1+t^{2})(1+4t^{2})}\ dt$ , where

$t=\tan x$

$\Rightarrow I=-\dfrac{1}{3}\displaystyle\int_{0}^{\infty}\left(\dfrac{1}{1+t^{2}}-\dfrac{4}{1+t^{2}}\right)dt=-\dfrac{1}{3}\left[\tan^{-1}t-2\tan^{-1}2t\right]_{0}^{\infty}$

$\Rightarrow =-\dfrac{1}{3}\left(\dfrac{\pi}{2}-\pi\right)=\dfrac{\pi}{6}$

$\cfrac { 7 }{ 5 } \left( 1+\cfrac { 1 }{ { 10 }^{ 2 } } +\cfrac { 1.3 }{ 1.2 } .\cfrac { 1 }{ { 10 }^{ 4 } } +\cfrac { 1.3.5 }{ 1.2.3 } .\cfrac { 1 }{ { 10 }^{ 6 } } +....\infty \right) =\sqrt { 2 }$

0%
True
0%
False

The sum

$2 \times 5 + 5 \times 9 + 8 \times 13 + \ldots 10$ term is

0%
$4500$
0%
$4555$
0%
$5454$
0%
None of these

Explanation

Solution -

General term =

$(3n-1)(4n+1)$

$= 12n^{2}+3n-4n-1$

$= 12n^{2}-4n+3n-1$

$=12n^{2}-n-1$

$\displaystyle \sum_{n=1}^{10}12n^{2}-n-1$

$= \dfrac{12n(n+1)(2n+1)}{6}- \dfrac{n(n+1)}{2}-n$

$=2\times 10\times 11\times 21-55-10$

$= 4620-55-10$

$= 4555$

B is correct

$1+\cfrac{1}{4}+\cfrac{1.3}{4.8}+\cfrac{1.3.5}{4.8.12}+.....\infty=\sqrt{2}$

0%
True
0%
False

A series is given as:

$4+7+10+13+16+.....$ Find the sum of the series up to

$10$ terms.

0%
$110$
0%
$125$
0%
$162$
0%
$175$

Explanation

The given series is

$4+7+10+13+16+...$

$a=4,d=3$

The sum of series is

$S_n=\dfrac n2[2a+(n-1)d]$

$=\dfrac {10}2[2(4)+9(3)]\\$

$=5(8+27)\\$

$=5\times 35\\$

$=175$

The unit's place digit in

$(1446)^{4n + 3}$ is

0%
$n$
0%
$0$
0%
$6$
0%
None of these

The sum of the infinite series,

${ 1 }^{ 2 }-\dfrac { { 2 }^{ 2 } }{ 5 } +\dfrac { { 3 }^{ 2 } }{ { 5 }^{ 2 } } -\dfrac { { 4 }^{ 2 } }{ { 5 }^{ 3 } } +\dfrac { { 5 }^{ 2 } }{ { 5 }^{ 4 } } -\dfrac { { 6 }^{ 2 } }{ { 5 }^{ 3 } } + ..........$ is

0%
$\dfrac { 1 }{ 2 }$
0%
$\dfrac { 25 }{ 24 }$
0%
$\dfrac { 25 }{ 54 }$
0%
$\dfrac { 125 }{ 252 }$

The sum of infinity of the series

$\dfrac{1}{1} + \dfrac{1}{1 + 2} + \dfrac{1}{1+2+3}+$ ______ is equal to:

0%
$2$
0%
$\cfrac{5}{2}$
0%
$3$
0%
None of these

Solve then inequality

$\dfrac {x-1}{x}\geq 2$

0%
$x\leq 1$
0%
$x\leq -1$
0%
$x\leq0$
0%
$x \in R$

Explanation

Given

$\dfrac {x-1}{x}\geq 2$

$x-1\geq 2x$

$x+1\leq 0$

$\implies x\leq -1$

$\implies x \in (-\infty ,-1]$

The sum of the series

$1+2.2+3.2^{2}+4.2^{3}+5.2^{4}+.+100.2^{99}$ is ?

0%
$99.2^{100} -1$
0%
$100.2^{100} +1$
0%
$99.2^[100}$
0%
$99.2^{100}+1$

Explanation

$S=1+2.2+{ 3.2 }^{ 2 }+{ 4.2 }^{ 3 }3+{ 5.2 }^{ 4 }+..+{ 100.2 }^{ 99 }$

$2S={ 1 }.{2}^{ 1 }+{ 2.2 }^{ 2 }+{ 3.2 }^{ 3 }+..+{ 99.2 }^{ 99 }+{ 100.2 }^{ 100 }$

subtracting the both,

$\left( 1-2 \right) S=1+{ 1.2 }^{ 1 }+{ { 1.2 } }^{ 2 }+{ { 1.2 } }^{ 3 }+..+{ 1.2 }^{ 99 }-{ 100.2 }^{ 100 }$

$-S=\left( 1+2+{ 2 }^{ 2 }+{ 2 }^{ 3 }+{ 2 }^{ 4 }+....+{ 2 }^{ 99 } \right) -{ 100.2 }^{ 100 }$

$-S=\left( \frac { 1\left( { 2 }^{ 100 }-1 \right) }{ 2-1 } \right) -{ 100.2 }^{ 100 }$

$-S={ 2 }^{ 100 }-1-{ 100.2 }^{ 100 }$ \\

$-S=-1-{ 99.2 }^{ 100 }$

$S=1+{ 99.2 }^{ 100 }$

${ S }_{ n }=\overset { n }{ \underset { r=1 }{ \Sigma } } { t }_{ r }=\dfrac { 1 }{ 6 } n\left( 2{ n }^{ 2 }+9n+13 \right)$ , then

$\overset { n }{ \underset { r=1 }{ \Sigma } } \sqrt { { t }_{ r } }$ equals ?

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

$a_n=n(n!)$ , then

$\displaystyle\sum^{100}_{r=1} a_r$ is equal?

0%
$101!$
0%
$100!-1$
0%
$101!-1$
0%
$101!+1$

Explanation

Given an=

$n(n!)$

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

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

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

$\therefore \displaystyle \sum_{r=1}^{100}a_{r}=a_{1}+a_{2}+a_{3}+.......+a_{100}$

$=((1+1)!-1!)+((2+1)!-2!)+....+((100+1)!-100!)$

$=2!-1!+3!-2!+.....+100!-99!+101!-100!$

$=101!-1$

$a_{1}=a_{2}=2,a_{n}=a_{n-1}-1(n > 2)$ then

$a_{5}$ is ?

0%
$1$
0%
$-1$
0%
$0`$
0%
$-2$

Explanation

Given that

${ a }_{ 1 }=2\quad and\quad { a }_{ 2 }=2$

use

${ a }_{ n }={ a }_{ n-1 }-1$

Putting n=3,

${ a }_{ 3 }={ a }_{ 3-1 }-1$

${ a }_{ 3 }={ a }_{ 2 }-1=2-1=1$

Putting n=4,

${ a }_{ 4 }={ a }_{ 4-1 }-1$

${ a }_{ 4 }={ a }_{ 3 }-1=1-1=0$

Putting n=5,

${ a }_{ 5 }={ a }_{ 5-1 }-1$

${ a }_{ 5 }={ a }_{ 4 }-1=0-1=-1$

Hence the five terms of sequence are

$2,2,1,0 and-1.$

The sum of the series

$1+\dfrac{1}{4\times 2!}+\dfrac{1}{16\times 4!}+\dfrac{1}{64\times 6!}+....\infty$ is?

0%
$\dfrac{e+1}{\sqrt{e}}$
0%
$\dfrac{e-1}{\sqrt{e}}$
0%
$\dfrac{e+1}{2\sqrt{e}}$
0%
$\dfrac{e-1}{2\sqrt{e}}$

Explanation

given series is

$1+\cfrac{1}{4\times 2!}+\cfrac{1}{16\times 4!}+\cfrac{1}{64\times 6!}+....\infty$

We know that

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

Let the series be denoted by

$S$

$S=1+\cfrac{1}{2!}.{ \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }+\cfrac{1}{4!}{ \left( \cfrac { 1 }{ 2 } \right) }^{ 4 }+\cfrac{1}{6!}{ \left( \cfrac { 1 }{ 2 } \right) }^{ 6 }+....\infty$

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

$2S=(1+\cfrac{1}{1!}(\cfrac{1}{2})+\cfrac{1}{2!}{ \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }+.... \infty)+(1-\cfrac{1}{1!}(\cfrac{1}{2})+\cfrac{1}{2!}{ \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }-\cfrac{1}{3!}{ \left( \cfrac { 1 }{ 2 } \right) }^{ 3 }+....\infty)$

${e}^{x}$ expansion

${e}^{1/2}=1+\cfrac{1}{1!}(\cfrac{1}{2})+\cfrac{1}{2!}{ \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }+....\infty)$

${e}^{-1/2}=1-\cfrac{1}{1!}(\cfrac{1}{2})+\cfrac{1}{2!}{ \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }+...\infty$

$2S={e}^{1/2}+{e}^{-1/2}$

$$S=\cfrac { \sqrt { e } +\cfrac { 1 }{ \sqrt { e } } }{ 2 } =\cfrac { e+1 }{ 2\sqrt { e } }

${x}_{1},{x}_{2},. . . . .,{x}_{n}$ are any real number and

$n$ is any positive integer, then ?

0%
$n\sum _{ i=1 }^{ n }{ { x }_{ i }^{ 2 }<{ \left( \sum _{ i=1 }^{ n }{ { x }_{ i } } \right) }^{ 2 } }$
0%
$n\sum _{ i=1 }^{ n }{ { x }_{ i }^{ 2 }\ge { \left( \sum _{ i=1 }^{ n }{ { x }_{ i } } \right) }^{ 2 } }$
0%
$n\sum _{ i=1 }^{ n }{ { x }_{ i }^{ 2 }\ge n{ \left( \sum _{ i=1 }^{ n }{ { x }_{ i } } \right) }^{ 2 } }$
0%
$None\ of\ the\ above$

Sum of first n terms of the series

$\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{{15}}{{16}} + ....$ is equal to

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

The

$nth$ term of the series

$4, 14, 30, 52, 80, 114, ...$ is

0%
$n^{2} + n + 2$
0%
$3n^{2} + n$
0%
$3n^{2} - 5n + 2$
0%
$(n + 1)^{2}$

The sum of

$(n+1)$ terms of

$\frac { 1 }{ 1 } +\frac { 1 }{ 1+2 } +\frac { 1 }{ 1+2+3 } +.......is$

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

Explanation

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

$\dfrac{1}{1}+\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{n+1}$

we find

$1^{st}$ term =

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

$2^{nd}$ term =

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

$3^{rd}$ term =

$\dfrac{2}{3}-\dfrac{2}{4} = \dfrac{1}{6}$

and so on

$n^{m}$ term

$= \dfrac{2}{n}-\dfrac{2}{n+1}$

$(n+1)^{m}$ term

$= \dfrac{2}{n+1}-\dfrac{2}{n+2}$

Now finding the sum of 4 terms;

$\dfrac{2}{1}-\dfrac{2}{2}+\dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+\dfrac{2}{4}-\dfrac{2}{5}$

we get,

$\dfrac{2}{1} -\dfrac{2}{5}$

So finding the sum of all

$n+1$ terms

we get,

$\dfrac{2}{1}-\dfrac{2}{n+2}= 2-\dfrac{2}{n+1}$

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

$\Rightarrow \dfrac{2n+2-2}{n+1}$

$\Rightarrow \dfrac{2n}{n+1}$

Hence, option (b) is correct.

1159090_1138889_ans_2ab17240a55c4277be7ddc075803b07d.jpg

The sum of the series

$\dfrac {5}{1 \cdot 2 \cdot 3}+\dfrac {7}{3 \cdot 4 \cdot 5}+\dfrac {9}{5 \cdot 6 \cdot 7}+....$ is

0%
$\log \dfrac {8}{e}$
0%
$\log \dfrac {e}{8}$
0%
$\log2$
0%
$\log \dfrac {1}{2}$

If the sum of the series

$1+\dfrac{3}{x}+\dfrac{9}{x^{2}}+\dfrac{27}{x^{3}}+....$ to

$\infty$ is a finite number then

0%
$x>3$
0%
$x<-3$
0%
$x<-3\ or\ x>3$
0%
$None\ of\ these$

Explanation

Series: $1+\dfrac{3}{{x}^{1}}+\dfrac{9}{{x}^{2}}+\dfrac{27}{{x}^{3}}+……\infty$

This is a $G.P$

With $Q A=1$

$r=\dfrac{3}{x}$

${ s }_{ n }=\dfrac { A }{ 1-r } =\dfrac { 1 }{ 1-\dfrac { 3 }{ x } } \quad \left\{ \because r<1 \right\}$

${s}_{n}=\dfrac{1\left(x\right)}{x-3}$ ...... 1

Also ${s}_{n}=\dfrac{A}{r-1}=\dfrac{1}{3.1}=\dfrac{x}{3-x}$ ......2

$\left\{ \because r\quad >\quad 1 \right\}$

From $1$ and $2$

We infer:

${s}_{n}=\dfrac{1\left(-x\right)}{x-3}$ is true for

$\left\{ 3\quad <\quad x \right\}$

And

${s}_{n}=\dfrac{x}{3-x}$ is true then

$\left\{ x\quad <\quad 3 \right\}$

Option $x\left\{ x<-3\& x>3 \right\}$

The sum of the series

$1^{3}+3^{3}+5^{3}+....$ upto

$20$ terms is

0%
$319600$
0%
$321760$
0%
$306000$
0%
$347500$

The value of

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

$\dfrac{p}{q}$ , where

$p$ and

$q$ are relatively prime natural numbers. Then the value of

$(p^2+q^2)$ is equal to

0%
$4$
0%
$9$
0%
$10$
0%
$13$

Explanation

Given,

$\displaystyle S=\sum_{n=1}^{\infty }\frac{1}{(3n-2)(3n+1)}= \frac{1}{3}\sum_{n=1}^{\infty }[\frac{1}{3n-2}-\frac{1}{3n+1}]$

$S=\dfrac{1}{3}[(1-\dfrac{1}{4})+(\dfrac{1}{4}-\dfrac{1}{7})+(\dfrac{1}{7}-\dfrac{1}{10})+....]$

$S= \dfrac{1}{3}(1-0)= \dfrac{1}{3}$

Comparing with

$S= \dfrac{p}{q}\Rightarrow p=1\, q=3$

$p^{2}+q^{2}=1^{2}+(3)^{2}=10$

$\therefore$ Option C is correct

1136517_1141198_ans_01cf6dbc3c6c4d338049fc19e8c19967.jpg

Select the missing number from the given responses ?

0%
$15$
0%
$10$
0%
$12$
0%
$17$

$abc=1$ , then the value of

$\dfrac{1}{1+a+b^{-1}}+\dfrac{1}{1+b+c^{-1}}+\dfrac{1}{1+c+a^{-1}}$ is

0%
$abc$
0%
$a$
0%
$1$
0%
$3$

Explanation

To find

$\rightarrow$

$\dfrac{1}{1+a+b^{-1}}+\dfrac{1}{1+b+c^{-1}}+\dfrac{1}{1+c+a^{-1}}$

$abc=1 \rightarrow a=\dfrac{1}{bc}$

$\dfrac{1}{1+a+b^{-1}}+\dfrac{1}{1+b+c^{-1}}+\dfrac{1}{1+c+a^{-1}}=$

$= \dfrac{1}{1+\dfrac{1}{bc}+b^{-1}}+\dfrac{1}{1+b+\dfrac{1}{c}}+\dfrac{1}{1+c+bc}$

$\Rightarrow \dfrac{bc}{1+c+bc}+\dfrac{c}{1+c+bc}+\dfrac{1}{bc+c+1}=1$

The sum

$1 + 3 + 7 + 15 + 31 + \ldots$ to

$n$ term is

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

Explanation

Given series is 1+3+5+7 so on upto n

nth term of this series is

${2}^{n}-1$

so,this series can be expressed as ({2}^{1}-1)+({2}^{2}-1)+({2}^{3}-1)so on upto ({2}^{n}-1)

$\rightarrow({1}^{1}+{2}^{2}+{2}^{3}+---{2}^{n})-n$

$\rightarrow\dfrac{2({2}^{n}-1)}{2-1}-n$

$\rightarrow{2}^{n+1}-2-n$

Find:

$6,25,62,123,(?),341$

0%
$216$
0%
$214$
0%
$215$
0%
$218$

Explanation

$6=2^{3}-2=6$

$25=3^{3}-2=25$

$62=4^{3}-2=62$

$123=5^{3}-2=123$

$214=6^{3}-2=214$

$341=7^3-2=341$

Missing term is

$214$

Sum of

$n$ terms of the series

$8+88+888+.$ equals

0%
$\dfrac {8}{18}[10^{n+1}-9n-10]$
0%
$\dfrac {8}{18}[10^{n}-9n-10]$
0%
$\dfrac {8}{18}[10^{n+1}+9n-10]$
0%
$None\ of\ these$

Explanation

series

$= 8+88+888+...$

$\dfrac{a_2}{a_1} = \dfrac{88}{8} = 11$

and

$\dfrac{a_3}{a_2} = \dfrac{888}{88} = 51$

$\therefore \dfrac{a_2}{a_1} \neq \dfrac{a_3}{a_2}.$ This is not a G.P.

But,

$S = 8+88+888+..$

$= 8 (1+11+111+...)$

$\dfrac{8}{9} (9(1+11+111+..)$

$\dfrac{8}{9}(9+99+999+...)$

$\dfrac{8}{9}((10-1)+(10^2-1)+(10^3-1)+...)$

$\Rightarrow \dfrac{8}{9}(10+10^2+10^3...)-(1+1+1+..)$

$\Rightarrow \dfrac{8}{9} (\dfrac{10(10^n-1)-n}{10-1})$

$S\Rightarrow \dfrac{8}{9}(\dfrac{10(10^n-1)-9n}{9})$

$\dfrac{8}{9} (\dfrac{10^{n+1}-10-9n}{9})$ [G.P with A = 10, B = 10

$S = \dfrac{A(R^n-1)}{R-1}]$

$\Rightarrow \dfrac{8}{81}(10^{n+1}-9n-10)$

1142367_1143821_ans_a56ebd4b58294368803a7e15e5a757cd.jpg

CBSE Questions for Class 11 Engineering Maths Sequences And Series Quiz 8 - 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 8 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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