MCQ Questions for Class 11 Engineering Maths Sequences And Series Quiz 10

The sum of

$\displaystyle \sum^{\infty}_{n=1}\tan^{-1}\left(\dfrac {2}{n^{2}+n+4}\right)$ is equal to-

0%
$\tan^{-1}2$
0%
$\dfrac {\pi}{2}+\tan^{-1}2$
0%
$\dfrac {\pi}{2}-\tan^{-1}2$
0%
$\dfrac {\pi}{4}$

$496 : 204 : : 329 : ?$

0%
$90$
0%
$110$
0%
$115$
0%
$135$

${ sin }^{ 4 }\alpha .{ cos }^{ 2 }\alpha =\sum _{ r=0 }^{ 3 }{ { C }_{ r } } .cos(2r\alpha )$ then

${ C }_{ 0 }+{ C }_{ 1 }+{ C }_{ 2 }+{ C }_{ 3 }=$

0%
0
0%
3
0%
4
0%
7

$S_n=\dfrac{7}{4.1.2}+\dfrac{10}{4^2.2.3}+\dfrac{13}{4^3.3.4}+....$ then

$S_{\propto}$ is equal to?

0%
$\dfrac{5}{2}$
0%
$\dfrac{9}{8}$
0%
$\dfrac{3}{2}$
0%
$1$

Find the missing number.

0%
$74$
0%
$62$
0%
$91$
0%
$97$

Explanation

$\textbf{ Step 1: Observing the difference between the adjacent numbers}$

$\text{If we notice the adjacent numbers, we see } 4\times 2 -1 =7$

$\text{Similarly, } 7 \times 2-1=13$

$\text{Similarly, } 13\times2 -1=25$

$\text{Similarly, } 25\times2-1=49$

$\textbf{Step 2: Calculating the missing number}$

$\text{Thus, looking at the above observations, we can conclude the missing number will be:}$

$\Rightarrow 49\times 2-1=97$

$\textbf{Thus, the missing number is D 97}$

The sum of the following series

$1+6+\dfrac{9(1^2+2^2+3^2)}{7}+\dfrac{12(1^2+2^2+3^2+4^2)}{9}+\dfrac{15(1^2+2^2+....+5^2)}{11}+.....$ up to

$15$ terms is:

0%
$7820$
0%
$7830$
0%
$7520$
0%
$7510$

Explanation

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

$T_n=\dfrac{3n.\dfrac{n(n+1)(2n+1)}{6}}{2n+1}=\dfrac{n^2(n+1)}{2}$

$S_{15}=\dfrac{1}{2}\displaystyle \sum_{n=1}^{15}(n^3+n^2)=\dfrac{1}{2}\left[\left(\dfrac{15(15+1)}{2}\right)^2+\dfrac{15\times 16\times 31}{6}\right]$

$=7820$

The value of the series

$\dfrac{1}{3!}+\dfrac{2}{5!}+\dfrac{3}{7!}+.....$ to

$\infty$ is equal to

0%
$\dfrac{1}{2}e$
0%
$\dfrac{1}{2e}$
0%
$\dfrac{3}{2e}$
0%
$\dfrac{4}{5e}$

Explanation

$\dfrac{0}{1!}+\dfrac{1}{3!}+\dfrac{2}{5!}+\dfrac{3}{7!}+\dfrac{4}{9!}+...=S$

$S=\sum_{k=0}^{\infty}{\dfrac{k}{\left(2k+1\right)!}}$

$=\dfrac{1}{2}\sum_{k=0}^{\infty}{\dfrac{2k}{\left(2k+1\right)!}}$

$=\dfrac{1}{2}\sum_{k=0}^{\infty}{\dfrac{2k+1-1}{\left(2k+1\right)!}}$

$=\dfrac{1}{2}\left[\sum_{k=0}^{\infty}{\dfrac{2k+1}{\left(2k+1\right)!}}-\sum_{k=0}^{\infty}{\dfrac{1}{\left(2k+1\right)!}}\right]$

$=\dfrac{1}{2}\left[\sum_{k=0}^{\infty}{\dfrac{1}{\left(2k\right)!}}-\sum_{k=0}^{\infty}{\dfrac{1}{\left(2k+1\right)!}}\right]$

$=\dfrac{1}{2}\sum_{k=0}^{\infty}{\dfrac{1}{\left(2k\right)!}}-\dfrac{1}{2}\sum_{k=0}^{\infty}{\dfrac{1}{\left(2k+1\right)!}}$

Now,

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

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

$\Rightarrow\,e+{e}^{-1}=2\left[1+\dfrac{1}{2!}+\dfrac{1}{4!}+...\right]=2\sum_{k=1}^{\infty}{\dfrac{1}{\left(2k\right)!}}$

$\Rightarrow\,\sum_{k=1}^{\infty}{\dfrac{1}{\left(2k\right)!}}=\dfrac{1}{2}\left(e+{e}^{-1}\right)$

$e-{e}^{-1}=2\left[1+\dfrac{1}{3!}+\dfrac{1}{5!}+...\right]=2\sum_{k=1}^{\infty}{\dfrac{1}{\left(2k+1\right)!}}$

$\Rightarrow\,\sum_{k=1}^{\infty}{\dfrac{1}{\left(2k+1\right)!}}=\dfrac{1}{2}\left(e-{e}^{-1}\right)$

$\therefore\,S=\dfrac{1}{2}\left(\dfrac{e+{e}^{-1}}{2}\right)-\dfrac{1}{2}\left(\dfrac{e-{e}^{-1}}{2}\right)$

$=\dfrac{1}{4}\left(e+{e}^{-1}-e+{e}^{-1}\right)$

$=\dfrac{1}{4}\left(2{e}^{-1}\right)$

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

The sum of the series

$\dfrac { 9 }{ 5^{ { 2 } }\cdot 2.1 } +\dfrac { 13 }{ 5^{ { 3 } }\cdot 3.2 } +\dfrac { 17 }{ 5^{ { 4 } }\cdot 4.3 } ..........$ Infinite terms

0%
$\dfrac { 1 } { 25 }$
0%
$\dfrac { 1 } { 5 }$
0%
Not a finite number
0%
$\dfrac { 9 } { 5 }$

Explanation

$\frac{9}{5^{2} \cdot 2 \cdot 1}+\frac{13}{5^{3} \cdot 3 \cdot 2}+\frac{17}{5^{4} \cdot 4 \cdot 3}................ \quad \infty \text { term }$

The general term of the above series,

for,

$r \geqslant 2$

General term of the numerator

$T_{r}^{\prime}=4 r+1$

General term of the denominater

$T_{r^{\prime \prime}}=\frac{1}{r(r-1)}$

Then

$T_{r}=\frac{4 r+1}{r(r-1) \cdot 5^{r}} \quad$ where,

$r \geqslant 2$

$=\frac{5 r-(r-1)}{5^{r}r(r-1)}$

$=\frac{5 r}{5^{r} \cdot(r(r-1)}-\frac{r-1}{r \cdot(r-1)}$

$\Rightarrow \frac{1}{5^{r-1}(r-1)}-\frac{1}{5^{r} \cdot r}$

Now, putting the

$r \geqslant 2$ in above equation

$T_{2}=\frac{1}{5 \cdot 1}-\frac{1}{5^{2} \cdot 2}$

$T_{3}=\frac{1}{5^{2} \cdot 2}-\frac{1}{5^{3} \cdot 3}$

$\because \text { every last term is gettimg cancelled }$

$\therefore \text { o } \quad \sum_{r=2}^{\infty} t_{r}=\frac{1}{5}$

Option B

The sum to infinity of the series

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

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

Explanation

Let

$S = 1 + \cfrac{2}{3} + \cfrac{6}{{3}^{2}} + ..... \quad \quad ..... \left( 1 \right)$

$\cfrac{S}{3} = \cfrac{1}{3} + \cfrac{2}{{3}^{2}} + \cfrac{6}{{3}^{3}} + ..... \quad \quad ..... \left( 2 \right)$

Subtracting equation

$\left( 2 \right)$ from

$\left( 1 \right)$ , we have

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

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

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

$S = 2 + \cfrac{2}{3} + \cfrac{2}{{3}^{2}} + \cfrac{2}{{3}^{3}} + .....$

Since the above series is in G.P. of infinite terms with common ration

$\cfrac{1}{3}$ .

$\therefore S = \cfrac{2}{1 - \cfrac{1}{3}} \\ \Rightarrow S = \cfrac{2 \times 3}{2} = 3$

Hence, this is the answer.

$x_{1},x_{2},....$ are in H.P and

$x_{1},2,x_{20}$ are in G.P, then

$\displaystyle \sum _{ 1 }^{ 19 }{ x_{r} .x_{r+1}}=$

0%
$76$
0%
$80$
0%
$84$
0%
$70$

The value of

$\dfrac { 1 }{ 4 } \tan\dfrac { \pi }{ 8 } +\dfrac { 1 }{ 8 } \tan\dfrac { \pi }{ 16 } +\dfrac { 1 }{ 16 } \tan\dfrac { \pi }{ 32 } +......\infty$ terms is equal to -

0%
$\dfrac { 5 }{ \pi } -\cfrac { 1 }{ 2 }$
0%
$\cfrac { 3 }{ \pi } +\cfrac { 1 }{ 2 }$
0%
$\cfrac { 2}{ \pi } -\cfrac { 1 }{ 2 }$
0%
$\cfrac { 4 }{ \pi } -\cfrac { 1 }{ 4 }$

Explanation

Let

$S=\dfrac{1}{4}\tan \dfrac{\pi}{8}+\dfrac{1}{8}\tan\dfrac{\pi}{16}+\dfrac{1}{16}\tan\dfrac{\pi}{32}+.....\infty$

$=\dfrac{1}{2}\left\{\dfrac{1}{2}\tan \left(\dfrac{\pi/4}{2}\right)+\dfrac{1}{4}\tan\left(\dfrac{\pi/4}{4}\right)+\dfrac{1}{8}\tan\left(\dfrac{\pi/4}{8}\right)+.….\infty\right\}$

Let

$x=\pi/4$

$\therefore s=\dfrac{1}{2}\left\{\dfrac{1}{2}\tan\left(\dfrac{x}{2}\right)+\dfrac{1}{2^2}\tan\left(\dfrac{x}{2^2}\right)+\dfrac{1}{2^3}\tan\left(\dfrac{x}{2^3}\right)+....\infty\right\}$

$=\dfrac{1}{2}\displaystyle\sum^{\infty}_{n=1}\left(\dfrac{1}{2^n}\tan\dfrac{x}{2^n}\right)$

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

$\left[\because \displaystyle\sum^{\infty}_{n=1}\dfrac{1}{2^n}\tan \dfrac{x}{2^n}=\dfrac{1}{x}-\cot x\right]$

$=\dfrac{1}{2}\left(\dfrac{1}{\pi/4}-\cot\pi/4\right)$

$=\dfrac{1}{2}\times \dfrac{\pi}{4}-\dfrac{1}{2}\times \cot\pi/4$

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

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

Find the missing number

0%
74
0%
62
0%
91
0%
97

Explanation

$\textbf{Step 1: Observing the difference between the adjacent numbers}$

$\text{If we notice the adjacent numbers, we see } 4\times 2 -1 =7$

$\text{Similarly, } 7 \times 2-1=13$

$\text{Similarly, } 13\times2 -1=25$

$\text{Similarly, } 25\times2-1=49$

$\textbf{Step 2: Calculating the missing number}$

$\text{Thus, looking at the above observations, we can conclude the missing number will be:}$

$\Rightarrow 49\times 2-1=97$

$\textbf{Thus, the missing number is D .}$

$an =\displaystyle \sum_{r =0}^{n} \dfrac{1}{^{n}C_{r}}$ then

$\displaystyle \sum_{r=0}^{n} \dfrac{r}{^{n}C_{r}}$ equals

0%
$(n-1)a_{n}$
0%
$na_{n}$
0%
$\dfrac{1}{2}na_{n}$
0%
$None\ of\ these$

Explanation

$\begin{aligned} \text { Solution- Given that } \\ \qquad a_{n} &=\sum_{r=0}^{n} \frac{1}{n_{n}} \\ S &=\sum_{r=0}^{n} \frac{n}{{ }^{n} C_{\mu}} \\ &=\sum_{n=0}^{n} \frac{n-(n-\mu)}{{ }^{n} C_{n}}=\sum_{n=0}^{n} \frac{n}{{ }^{n} C_{\mu}}-\sum_{n=0}^{n} \frac{n-n}{{ }^{n} C_{\mu}} \end{aligned}$

As we know that

${ }^{n} C_{n}={ }^{n} C_{n-\mu}$

$\Rightarrow S=n \sum_{r=0}^{n} \frac{1}{n_{n}}-\sum_{n=0}^{n} \frac{n-n}{n_{n-n}}$

$\begin{array}{l} S=n \cdot(a n)-S \\ 2 S=n(a n) \\ S=\frac{1}{2} n a_{n} \\ \text { Hence, }(C) \text { is the correct option. } \end{array}$

Find the sum of first

$15$ terms of the sequence whose

${n}^{th}$ term is

$3+4n$ .

0%
$525$
0%
$425$
0%
$495$
0%
$None\,of\,thes$

Explanation

$n^{th}$ term = 3+4 n

$1^{st}$ term = 3+4(1)=7

$2^{nd}$ term = 3+4(2)=11

$3^{rd}$ term 3+4(3)=15

It is an AP with a=7, d=4

$S_{15}=\frac{15}{2}[2(7)+14(4)]$

$= 15[7+28]$

$= 525$

1184070_1337633_ans_b4d2fc976d0d4934a6c3a65dc922e6fc.jpg

If in a series

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

$\displaystyle\sum^{10}_{n=0}t_n$ is equal to?

0%
$1-\dfrac{1}{10!}$
0%
$1-\dfrac{1}{11!}$
0%
$1-\dfrac{1}{12!}$
0%
None of these

Explanation

$\begin{array}{l} \text { Solution - It is given that } \\ \qquad \begin{aligned} t_{n} &=\frac{n+1}{(n+2) !} \\ t n &=\frac{n+1+1-1}{(n+2) !}=\frac{(n+2)-1}{(n+2) !} \\ & \text { th }=\frac{n+2}{(n+2) !}-\frac{1}{(n+2) !} \end{aligned} \end{array}$

$t n=\frac{1}{(n+1) !}-\frac{1}{(n+2) !}$

$\sum_{n=0}^{10} t_{n}=\sum_{n=0}^{10}\left(\frac{1}{(n+1) !}-\frac{1}{(n+2) !}\right)$

$\begin{aligned} &=\frac{1}{1}\left(-\frac{1}{2 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{3 !}-\frac{1}{4 !} \cdots \cdots-\frac{1}{11 !}+\frac{1}{11 !}\right)-\frac{1}{12 !} \\ &=1-\frac{1}{12 !} \\ \text { Hence, } &(\mathrm{C}) \text { is the correct option. } \end{aligned}$

Let

$\sum _{ r=1 }^{ n }{ { r }^{ 4 } } =f\left( n \right) ,$ then

$\sum _{ r=1 }^{ n }{ { \left( 2r-1 \right) }^{ 4 } }$ is equal to

0%
$f\left( 2n \right) -16f\left( n \right)$
0%
$f\left( 2n \right) +7f\left( n \right)$
0%
$f\left( 2n+1 \right) 8f\left( n \right)$
0%
none of these

If the sum of the first 15 terms of the series

$\lgroup \frac{3}{4} \rgroup^3 + \lgroup 1\frac{1}{2} \rgroup^3 + \lgroup 2\frac{1}{4} \rgroup^3 + 3^3 + \lgroup 3\frac{3}{4} \rgroup^3 + ...........$ is equal to 225 k. then k is equal to:

0%
9
0%
27
0%
108
0%
54

Explanation

$\lgroup \dfrac{3}{4} \rgroup^3 + \lgroup \dfrac{6}{4} \rgroup^3 + \lgroup \dfrac{9}{4} \rgroup^3 + \lgroup \dfrac{12}{4} \rgroup^3 + .............. 15$ term

$S=\displaystyle\sum_{r=1}^{15}\left [\dfrac{3r}{4}\right ]^3=\dfrac{27}{64}\sum r^3$

$=\displaystyle\dfrac{27}{64} \sum_{r = 1}^{15} r^3$

$= \dfrac{27}{64}. \Big[ \dfrac{15(15+1)}{2} \Big ]^2$

$= 27\times 225$ (given in question)

$K = 27$

$C_1 + 2C_2 + 3C_3 + ...... + nC_n$ is equal to

0%
$2^{n-1}$
0%
$2^{n +1}$
0%
$n. 2^{n -1}$
0%
$n. 2^{n +1}$

Explanation

Given that,

${{C}_{1}}+2.{{C}_{2}}+3.{{C}_{3}}+.............n{{C}_{n}}$

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

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

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

Put, n-1=N

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

$=n\left( {}^{N}{{C}_{0}}+{}^{N}{{C}_{1}}+{}^{N}{{C}_{2}}...............{}^{N}{{C}_{N}} \right)$

$=n{{.2}^{N}}$

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

Hence, this is the answer.

insert the missing number: 5,8,12,17,23,___,38

0%
$29$
0%
$30$
0%
$32$
0%
$25$

Explanation

$5,8,12,17,23,......38$

$5+3=8;8+4=12,12+5=17,17+6=23$

$23+7=30,30+8=30$

$\therefore 30$

The sum of the series

$1+\dfrac{\log_e x}{1!}+\dfrac{(\log_e x)^2}{2!}+........$ is

0%
$x$
0%
$x^2$
0%
$x^3$
0%
none of these

Explanation

We've,

$1+\dfrac{\log_e x}{1!}+\dfrac{(\log_e x)^2}{2!}+........$

$=e^{\log_ex}$ [ Since

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

$=x$ .

If x<1, then

$\displaystyle \frac { 1 }{ 1+x } +\frac { 2x }{ 1+{ x }^{ 2 } } +\frac { { 4x }^{ 3 } }{ 1+{ x }^{ 4 } } +.........\infty =$

0%
$x$
0%
$\frac { 1 }{ 1+x }$
0%
$\frac { 1 }{ 1-x }$
0%
$\frac { 1 }{ x }$

The sum of the series 1+2(1 +1/n)+ 3

${ \left( 1+1/n \right) }^{ 2 }+....\infty \quad is\quad given\quad by$

0%
${ n }^{ 2 }\quad +\quad 1$
0%
$n\left( n+1 \right)$
0%
$n{ (1+1/n) }^{ 2 }$
0%
${ n }^{ 2 }$

Explanation

Given,

$S_m=1+2\left ( 1+\dfrac{1}{n} \right )+.........+\infty$

$\Rightarrow S_m\left ( 1+\dfrac{1}{n} \right )=\left ( 1+\dfrac{1}{n} \right )+2\left ( 1+\dfrac{1}{n} \right )^2+.....+\infty$

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

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

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

$S_m=n^2$

The sum of the series

$1 + \frac { 1 + 2 } { 2 } + \frac { 1 + 2 + 3 } { 3 } + \ldots$ to n terms is

0%
$\frac { n ( n + 1 ) } { 2 }$
0%
$\frac { n ( n + 3 ) } { 4 }$
0%
$\frac { n ( n + 1 ) + n } { 4 }$
0%
$\frac { n ( n - 1 ) } { 2 }$

$\sum _{ k=2 }^{ n }{ cos{ }^{ -1 }(\frac { 1+\sqrt { (k-1)(k+2)(k+1)k } }{ k(k+1) } } )=\frac { 120\pi }{ \lambda }$ ,then

0%
number of even divisors of $\lambda$ is 24
0%
Sum of proper divisors of $\lambda$ is 2418.
0%
Sum of proper divisors of $\lambda$ is 197.
0%
number of ways in which $\lambda$ can be expressed as product of two co-prime factors is 4

What is the next number in the series

$2,12,36,80,150?$

0%
$194$
0%
$210$
0%
$252$
0%
$258$

Explanation

$12 - 2 = 10$ .

$36 - 12 = 24$ .

$80 - 36 = 44$ .

$150 - 80 = 70$ .

Now,

$24 - 10 = 14$ .

$44 - 24 = 20$ .

$70 - 44 = 26$ .

Then next difference of the difference will be

$26 + 6 = 32$ .

Hence answer is

$150 + 70 + 32 = 252$ .

The sum of the first n terms of the series

${ 1 }^{ 2 }+2.{ 2 }^{ 2 }+{ 3 }^{ 2 }+2.{ 4 }^{ 2 }+{ 5 }^{ 2 }+2.{ 6 }^{ 2 }+...$ 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%
${ \left[ \dfrac { n\left( n+1 \right) }{ 2 } \right] }^{ 2 }$

Value of

$1+\dfrac{1^3+2^3}{1+2}+\dfrac{1^3+2^3+3^3}{1+2+3}+...+\dfrac{1^3+2^3+......15^3}{1+2+.....+15}-\dfrac{1}{2}(1+2+.....+15)$ is?

0%
$840$
0%
$720$
0%
$680$
0%
$620$

Explanation

$\displaystyle\sum^{15}_{r=1}\dfrac{\left(\dfrac{(n)(n+1)}{2}\right)^2}{\left(\dfrac{(n)(n+1)}{2}\right)}-\dfrac{1}{2}\left(\dfrac{15\times 16}{2}\right)=\displaystyle\sum^{15}_{r=1}\left(\dfrac{n^2}{2}+\dfrac{n}{2}\right)-60=620+60-60=620$ .

Find the sum of the following series to n terms:

$1\times 2+2\times 3+3\times 4+4\times 5+...$

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

Explanation

Let

$T_n$ be the

$n^{th}$ term of the given series, Then

$T_n=(n^{th}\ term\ of\ 1,2,3....)\times (n^{th}\ term\ of\ 2,3,4....)$

$=[1+(n-1)\times 1]. [2+(n-1)\times 1]$

$=[1+n-1][2+n-1]$

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

Let

$S_n$ denotes sum of

$n$ term of the given series

$S_n= \displaystyle \sum_{n=1}^{n}T_n =\displaystyle \sum_{n=1}^{n} (n^2 +n)=\displaystyle \sum_{n=1}^{n}n^2 +\displaystyle \sum_{n=1}^{n}n$

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

$=\dfrac {n(n+1)(2n+4)}{6}=\dfrac {n(n+1)\times 2(n+2)}{6}$

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

Find the sum of the following series to n terms:

$1+(1+2)+(1+2+3)+(1+2+3+4)+...$

0%
$\dfrac{n}{6}(n+1)(n+2)$
0%
$\dfrac{n}{6}(n-1)(n-2)$
0%
$\dfrac{n}{6}(n+1)(n-1)$
0%
None of these

Explanation

Let

$T_n$ be the

$n^{th}$ term of the given series,

Then,

$T_n =1+2+3+.....+n$

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

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

$=\dfrac {n^2}{2}+\dfrac {n}{2}$

Let

$S_n$ denoted sum of

$n$ terms of the given series

$S_n =\displaystyle \sum_{K=1}^nT_K=\displaystyle \sum_{K=1}^n \left [\dfrac {K^2}{2}+K/2\right]$

$S_n =\dfrac {1}{2} \displaystyle \sum_{K=1}^nK^2 +\dfrac {1}{2}\displaystyle \sum_{K=1}^nK$

$S_n=\dfrac {1}{2} \left [\dfrac {n(n+1) (2n+1)}{6} \right] +\dfrac {1}{2} \left [\dfrac {n(n+1)}{2}\right]$

$S_n=\dfrac {n(n+1)(2n+1)+3n(n+1)}{12}$

$S_n =\dfrac {n(n+1)[2n+4]}{12}$

$S_n=\dfrac {n(n+1)}{12}\times 2(n+2)$

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

5,5,5,... forms a ______

0%
Series
0%
Sequence
0%
Progression
0%
None of the above

Whole numbers from a sequence/progression, as they are formed by the fixed rule of adding 1 to previous whole number.

0%
True
0%
False

State true/false:

$3,4,5,6,7,......$ forms a progression. As they are formed by the rule

$n+1$

0%
True
0%
False

Find the sum of Arithmetic means of

$3 , 9$ and

$12 , 8$ .

0%
$6$
0%
$9$
0%
$20$
0%
$16$

The arithmetic mean of

$4$ and

$14$ is

0%
$9$
0%
$18$
0%
$10$
0%
$4$

If AM of two numbers a and 17 isThen a is ___

0%
15
0%
13
0%
17
0%
None of the above

Insert an arithmetic mean between

$7$ and

$21$

0%
$10$
0%
$14$
0%
$28$
0%
$30$

Arithmetic mean of two numbers a+d and a-d is

0%
$a$
0%
$b$
0%
Cannot be determined
0%
None of the above

Sum of the series

$S=1^2-2^2+3^2-4^2+.....-2002^2+2003^2$ is

0%
$2007006$
0%
$1005004$
0%
$2000506$
0%
$1005040$

Explanation

$S = {1}^{2} - {2}^{2} + {3}^{2} - {4}^{2} + ...... - {2002}^{2} + {2003}^{2}$

$\displaystyle \Rightarrow S = \sum_{k = 1}^{2003}{{\left( -1 \right)}^{k + 1}{k}^{2}}$

$\displaystyle \Rightarrow S = \sum_{k = 1}^{1002}{{\left( 2k - 1 \right)}^{2}} - \sum_{k = 1}^{1001}{{\left( 2k \right)}^{2}}$

$\displaystyle \Rightarrow S = {\left[ \left( 2 \times 1002 \right) - 1 \right]}^{2} + \sum_{k = 1}^{1001}{\left[ {\left( 2k \right)}^{2} - {\left( 2k \right)}^{2} - 4k + 1 \right]}$

$\displaystyle \Rightarrow S = {\left( 2003 \right)}^{2} - \sum_{k = 1}^{1001}{\left( 4k - 1 \right)} = {\left( 2003 \right)}^{2} - 4 \times \cfrac{1001 \left( 1001 + 1 \right)}{2} + 1001 = 2007006$

The sum of

$n$ terms of the series

$1^2-2^2+3^2-4^2+5^2-6^2+...$ is

0%
$\dfrac{-n(n+1)}{2}$ if $n$ is even
0%
$\dfrac{n(n+1)}{2}$ if $n$ is odd
0%
$-n(n+1)$ if $n$ is even
0%
$\dfrac{n(n+1)(2n+1)}{6}$ if $n$ is odd

Explanation

If n is even

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

$n(n+1)(2n+1)/6-8(n/2(n/2+1)(n+1))/6$

$-n(n+1)/2$

if nis odd

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

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

$n(n+1)(2n+1)/6-8((n-1)/2\times(n+1)/2\times(n))/6$

$-2n^{2}$

$=n(n+1)/2$

$\displaystyle \frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+\ldots.n$ terms

$=$

0%
$\displaystyle \frac{\mathrm{n}(2\mathrm{n}^{2}+9\mathrm{n}+13)}{24}$
0%
$\displaystyle \frac{n(2n^{3}+9n+13)}{8}$
0%
$\displaystyle \frac{n(n^{2}+9n+13)}{24}$
0%
$\displaystyle \frac{n(n^{2}+9n+13)}{8}$

Explanation

General

$\displaystyle k^{th}$ term

$= \dfrac{\sum_{i=1}^{k}i^3}{\sum_{i=1}^{k}(2i-1)}$

$\displaystyle =\frac{(k+1)^2}{4}$

Final answer :

$\displaystyle \sum_{k=1}^n\dfrac{(k^2+2k+1)}{4}$

$\implies \displaystyle S_n = \dfrac{1}{4}\displaystyle \sum_{k=1}^n k^2+ 2k+1$

We know,

$\sum_{i=1}^{n} k =1 + 2 + 3 +… + n = \dfrac{n(n+1)}{2}$ (sum of first n natural numbers)

$\sum_{i=1}^{n} k^2 =1^2 + 2^2 + 3^2 +… + n^2=\dfrac{n(n+1)(2n+1)}{6}$ (sum of squares of the first n natural numbers)

$\therefore S_n =\displaystyle \frac{n(n+1)(2n+1)}{24}+\frac{n(n+1)}{4}+\frac{n}{4}$

$\displaystyle=n\frac{(2n^2+9n+13)}{24}$

Hence, option 'A' is correct.

Observe the following lists List I and List II

(A)

$\displaystyle \sum_{n=0}^{\infty}\frac{x^{n}(\log_{e}a)^{n}}{n!}$

$(1)\displaystyle \frac{e^{x}-e^{-x}}{2}$
(B)

$\displaystyle \sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}$

$(2) e^{-ax}$
(C)

$\displaystyle \sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)!}=$

$(3) a^{x}$
(D)

$\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n}.(ax)^{n}}{n!}$

$(4)\displaystyle \frac{a^{x}-a^{-x}}{2}$

$(5)\displaystyle \frac{e^{x}+e^{-x}}{2}$
The correct match of List I to List II is:

0%
A - 1, B - 4, C - 3, D - 5
0%
A - 4, B - 2, C - 1, D - 3
0%
A - 3, B - 5, C - 1, D - 2
0%
A - 2, B - 3, C - 5, D - 4

Explanation

(A) When we expand it, we will get

$1+\dfrac { { x }^{ }{ \log _{ e }{ a } }^{ } }{ 1 } + \dfrac { { x }^{ 2 }{( \log _{ e }{ a } })^{2 } }{ 2! }+....$ This is a general expression for the expansion of

${ a }^{ x }$

So, A - 3

(B) Expansion is given by

$1+\dfrac { { x }^{ 2 } }{ 2! } +\dfrac { { x }^{ 4 } }{ 4! } +\dfrac { { x }^{ 6 } }{ 6! } ..$ which is

$\dfrac { { e }^{ x }+{ e }^{ -x } }{ 2 }$

So, B - 5

$1+\dfrac { { x }^{ 3 } }{ 3! } +\dfrac { { x }^{ 5 } }{ 5! } +\dfrac { { x }^{ 7 } }{ 7! } ..$ and this is

$\dfrac { { e }^{ x }-{ e }^{ -x } }{ 2 }$

So, C - 1

(D) Expansion is given by

$1-\dfrac { ax }{ 1! } +\dfrac { (ax)^{ 2 } }{ 2! } -\dfrac {(ax)^{3}}{ 3! }....$ which is an expansion of

${ e }^{ -ax }$

So, D - 2

Sum the series

$1^3+3^3+5^3+..........$ to

$n$ terms is

0%
$n^2\left ( 2n^2-1 \right )$
0%
$n\left ( 2n^2-1 \right )$
0%
$n\left ( 2n^2+1 \right )$
0%
$n^2\left ( 2n^2+1 \right )$

Explanation

$n^{th}$ term of the series will be

$(2n-1)^3$

$1^3 +3^3 +5^3+.....+(2n-1)^3 = 1^3 +2^3 +3^3+4^3+.....(2n)^3 -(2^3 +4^3+6^3+......(2n)^3)$

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

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

$=4n^4+4n^3+n^2-2n^4-4n^3-2n^2$

$=2n^4-n^2$

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

$\displaystyle f(r)=1+\frac {1}{2}+\frac {1}{3}+.....+\frac {1}{r}$ and

$f(0)=0$ , then value of

$\displaystyle \sum_{r=1}^{n}(2r+1)f(r)$ is

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

Explanation

Since

$\displaystyle \sum_{r=1}^{n}(2r+1)f(r)=\sum_{r=1}^{n}(r^2+2r+1-r^2)f(r)=\sum_{r=1}^{n}\left \{(r+1)^2-r^2\right \}f(r)$

$\displaystyle =\sum_{r=1}^{n}\left \{(r+1)^2f(r)-(r+1)^2f(r+1)+(r+1)^2f(r+1)-r^2f(r)\right \}$

$\displaystyle =\sum_{r=1}^{n}(r+1)^2\left \{f(r)-f(r+1)\right \}+\sum_{r=1}^{n}\left \{(r+1)^2f(r+1)-r^2f(r)\right \}$

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

$\displaystyle =-\sum_{r=1}^{n}(r+1)+\left \{2^2f(2)+3^2f(3)+ ....+n^2f(n)\right \}+(n+1)^2f(n+1)-\left \{1^2f(1)+2^2f(2)+3^2f(3)+ .....+n^2f(n)\right \}$

$\displaystyle =-\sum_{r=1}^{n}r-\sum_{r=1}^{n}1+\left((n+1)^2f(n+1)-1^2f(1)\right)=-\frac {n(n+1)}{2}-n+(n+1)^2f(n+1)-f(1)$

$\displaystyle =(n+1)^2f(n+1)-\frac {n(n+3)}{2}-1 =(n+1)^2f(n+1)-\frac {(n^2+3n+2)}{2}$

$n$ is an odd integer greater than or equal to

$1$ , then the value of

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

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

Explanation

The given series is

$S = 1^3 -2^3 +3^3 -4^3 +5^3 -6^3+...-(n-1)^3+n^3$

$\because n$ is an odd integer.

This is the same as

$S = (1^3 +2^3 +3^3 +...+n^3)-2(2^3+4^3+6^3+...+(n-1)^3)$

Which can be written as

$S = (1^3 +2^3 +3^3 +...+n^3)-16\left(1^3 +2^3 +3^3 +...\left(\dfrac {n-1}{2}\right)^3\right)$

Which can now be solved using

$\displaystyle\sum r^3 =\dfrac {n^2(n+1)^2}{4}$

$\therefore$ The desired answer is

$S = \dfrac{n^2(n+1)^2}{4} - 16\dfrac{\left(\dfrac{n-1}{2}\right)^2\left(\dfrac{n+1}{2}\right)^2}{4} = \dfrac{(n+1)^2(2n-1)}{4}$

Hence, option A is the correct answer.

The sum to

$n$ terms of the series

$\displaystyle \frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+...$ is

0%
$\displaystyle \frac{6n}{n+1}$
0%
$\displaystyle \frac{9n}{n+1}$
0%
$\displaystyle \frac{12n}{n+1}$
0%
$\displaystyle \frac{3n}{n+1}$

Explanation

General term of the given sequence is

$T_n = \dfrac{2n+1}{\displaystyle\sum_1^nn^2} = \dfrac{2n+1}{\dfrac{n(n+1)(2n+1)}{6}} = 6\left(\dfrac1{n} - \dfrac1{n+1}\right)$

Now,

$\displaystyle S_n = \sum_1^nT_n = 6\sum_1^n\left(\dfrac1n-\dfrac1{n+1}\right)$

$S_n = 6\left(1 - \dfrac1{n+1}\right) = \dfrac{6n}{n+1}$

The sum of the first

$n$ terms of the series

$1^2+2\cdot2^2+3^2+2\cdot4^2+5^2+2\cdot6^2+.....$ is

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

$n$ is even.

When

$n$ is odd, then the sum is

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

Explanation

$n$ is odd, then

$n-1$ is even.

$(n-1)^{th}$ term is

$2(n-1)^2$ the

$n^{th}$ term is

$n^2$
Hence, using the given formula is valid if

$n$ is even, so substitute

$n-1$ in that formula in place of

$n$ and add

$n^2$
We get,

Sum

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

The sum of n terms of the series

$1 + (1 + a) + (1 + a + a^2) + (1 + a + a^2 + a^3) + ....$ is

0%
$\displaystyle \frac{n}{1+a} - \frac{1 - a^n}{(1 - a)^2}$
0%
$\displaystyle \frac{n}{1- a} + \frac{a (1 - a^n)}{(1 - a)^2}$
0%
$\displaystyle \frac{n}{1 - a} + \frac{a (1 + a^n)}{(1 - a)^2}$
0%
none of the above

ABCD is a square of length a,

$a\epsilon N$ , a>Let

$L_{1}$ ,

$L_{2}$ ,

$L_{3}$ ,... be points on BC such that

$BL_{1}=L_{1}L_{2}=L_{2}L_{3}=...=1$ and

$M_{1}$ ,

$M_{2}$ ,

$M_{3}$ ,... be points on CD such that

$CM_{1}=M_{1}M_{2}=M_{2}M_{3}=...=1$ . Then

$\sum_{n=1}^{a-1}\left ( A{L_{n}}^{2}+L_{n}{M_{n}}^{2} \right )$ is equal to

0%
$\dfrac{1}{2}a\left ( a-1 \right )^{2}$
0%
$\dfrac{1}{2}a\left ( a-1 \right )\left ( 4a-1 \right )$
0%
$\dfrac{1}{2}a\left ( a-1 \right )\left ( 2a-1 \right )\left ( 4a-1 \right )$
0%
none of these

Explanation

$A{ { L }_{ n } }^{ 2 }={ a }^{ 2 }+{ n }^{ 2 }\\ { { L }_{ n }{ M }_{ n } }^{ 2 }={ (a-n) }^{ 2 }+{ n }^{ 2 }\\ A{ { L }_{ n } }^{ 2 }+{ { L }_{ n }{ M }_{ n } }^{ 2 }=2{ a }^{ 2 }+{ 3n }^{ 2 }-2na$
Thus, the summation is:

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

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

$\displaystyle \underset{n\rightarrow \infty}{ \lim} \sum_{r=1}^{n}\frac {1}{t_{r}}$
is equal to:

0%
$\displaystyle \frac {1}{8}$
0%
$\displaystyle \frac {1}{4}$
0%
$\displaystyle \frac {1}{2}$
0%
$1$

Let x, y, z be three positive prime numbers. The progression in which

$\sqrt{x}$ ,

$\sqrt{y}$ ,

$\sqrt{z}$ can be three terms (not necessarily consecutive) is

0%
AP
0%
GP
0%
HP
0%
none of these

Explanation

$x,y,z$ are primes

$\Rightarrow \sqrt{x},\sqrt{y},\sqrt{z}$ are irrationals

Let

$\sqrt{x}=a+pd; \sqrt{y}=a+qd; \sqrt{z}=a+rd$

Now,

$(\sqrt{y}-\sqrt{x})^{2} = ((q-p)d)^{2}$

$\Rightarrow x+y-2\sqrt{xy}=((q-p)d)^{2}$

LHS is irrational whereas RHS is rational

Not Possible

Hence, Not in AP

Let

$\sqrt{x}=ar^{p}; \sqrt{y}=ar^{q}; \sqrt{z}=ar^{s}$

$\therefore \dfrac{\sqrt{y}}{\sqrt{x}} = r^{q-p}$

$\Rightarrow y= x(r^{2(q-p)})$

$\Rightarrow y$ has two factors

$x$ and

$r^{2(q-p)}$

Not possible as

$y$ is prime

Hence, not in GP

Let

$\dfrac{1}{\sqrt{x}}=a+pd; \dfrac{1}{\sqrt{y}}=a+qd; \dfrac{1}{\sqrt{z}}=a+rd$

$\dfrac{1}{\sqrt{y}}-\dfrac{1}{\sqrt{x}} = (q-p)d$

$\Rightarrow \dfrac{(\sqrt{y}-\sqrt{x})^{2}}{yx} = ((p-q)d)^{2}$

$x+y-2\sqrt{yx} = xy((p-q)d)^{2}$

LHS = irrational and RHS = rational

Not possible. Hence, Not in HP

CBSE Questions for Class 11 Engineering Maths Sequences And Series Quiz 10 - 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 10 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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