Class 11 Commerce Maths - Sequences And Series

Find the following sum.
$\sum\limits_{r = 1}^n {\left( {6{r^2} - 2r + 6} \right)}$

$\begin{array}{l}\sum\limits_{r = 1}^n {\left( {6{r^2} - 2r + 6} \right)} \\ = 6\sum {{r^2}} - 2\sum r + 6\sum 1 \\ = 6\dfrac{{n\left( {n - 1} \right)\left( {2n + 1} \right)}}{6} - 2\dfrac{{n\left( {n + 1} \right)}}{2} + 6n\\ = n\left[ {\left( {n + 1} \right)\left( {2n + 1} \right) - \left( {n + 1} \right) + 6} \right]\\ = n\left[ {2{n^2} + 3n + 1 - n - 1 + 6} \right]\\ = n\left( {2{n^2} + 2n + 6} \right)\\ = 2n\left( {{n^2} + n + 3} \right)\end{array}$

$1.2 + 2.3x + 3.4x^{2} + ....\infty (|x| < 1)$ .

1521166_1153875_ans_961d610b0fb240ef9232e8dc83ba7b76.jpg

Write the general term for the given series $\frac{1}{(n-1)},\frac{1}{3(n-3)},\frac{1}{5(n-5)},....$

General term can be observed as

$\dfrac{1}{(2r-1)(n-(2r-1))}$ for r to be some natural number

Solve:

$\displaystyle\sum\limits_{r = 1}^n {\left( {6{r^2} - 2r + 6} \right)}$

$\displaystyle\int^n_{r=}(6r^2-2r+6)$

$\Rightarrow 2\displaystyle\sum^n_{r=1}[3r^2-r+3]$

$\Rightarrow 2\left[3\displaystyle\sum^n_{r=1}r^2-\displaystyle\sum^n_{r=1}r+3\displaystyle\sum^n_{r=1}(1)\right]$

$\Rightarrow 2\left[3\left[\dfrac{n(n+1)(2n+1)}{6}\right]-\dfrac{n(n+1)}{2}+3n\right]$

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

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

$\Rightarrow 2\left[n^2(n+1)+3n\right]$

$\Rightarrow 2n[n^2+n+3]$

$\therefore \displaystyle\sum^n_{r=1}6r^2-2r+6$

$=2n[n^2+n+3]$ .

1123619_1197135_ans_0aa92c313fc44c12ae6cb47f1ce630bd.jpeg

The product of two consecutive positive numbers is $182$ . Find the numbers.

A] Let two numbers be x , x + 1

given , x (x + 1) = 182

$= x^{2}+x-182 = 0$

$= x^{2}+14x-13x-182 = 0$

$= x (x + 14) - 13 (x + 14) = 0$

= (x + 14) (x - 13) = 0

as x is positive

so x = 13

Two numbers are 13 , 14

1177886_1292279_ans_902baba0409e417a918a156e805b95a8.jpg

Evaluate $1+i^{2}+i^{4}+i^{6}+...+i^{2n}$ .

$1+i^{2}+i^{4}+i^{6}+ ...+i^{2n}$

If n is odd.

$\Rightarrow 1-1+1-1+1-1=0$

If n is even

$\Rightarrow 1-1+1 ...+1=1$

What is the sum of first 20 odd natural numbers?

$\begin{array}{l} The\, first\, 20\, odd\, natural\, number\, is\, given\, by \\ 1,3,5,7,9,.....37,39. \\ here, \\ a=1 \\ d=2 \\ and \\ n=20 \\ Now, \\ { S_{ n } }=\dfrac { n }{ 2 } \left[ { 2{ a_{ 1 } }+\left( { n-1 } \right) d } \right] \\ { S_{ 20 } }=\dfrac { { 20 } }{ 2 } \left[ { 2\left( 1 \right) +\left( { 20-1 } \right) 2 } \right] \\ =10\left[ { 2+\left( { 19 } \right) 2 } \right] \\ =10\left[ { 2+38 } \right] \\ =10\left[ { 400 } \right] \\ =400 \\ Hence,\, the\, sum\, of\, first\, 20\, odd\, natural\, number\, is\, 400. \end{array}$

Sum the following series
$1 + \dfrac {3}{4} + \dfrac {7}{16} + \dfrac {15}{64} + \dfrac {31}{256} + ....$ to infinity.

We have

$S=1+\dfrac{3}{4}+\dfrac{7}{16}+\dfrac{15}{64}+\dfrac{31}{256}\dots\dots(1)$

Multiplying

$S$ by

$4$ , we get

$4S=4+3+\dfrac{7}{4}+\dfrac{15}{16}+\dfrac{31}{64}\dots\dots(2)$

$(2)-(1)$

$\Rightarrow 3S=6+1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}\dots\dots=6+2=8$

$\Rightarrow S=\dfrac{8}{3}$

Find value of the sum $\displaystyle \frac{1}{sin45^o sin 46^o} + \frac{1}{sin47^o sin 48^o} + \frac{1}{sin49^o sin 50^o} + ....... + \frac{1}{sin133^o sin 134^o}$

$\dfrac {1}{\sin 45 \sin 46}+\dfrac {1}{\sin 46 \sin 47}+....+\dfrac {1}{\sin 133 \sin 134}$

$\Rightarrow \$ Multiply and divide by

$\sin 1$ we get

$=\dfrac {1}{\sin 1} \left [\dfrac {\sin (46-45)}{\sin 45 \sin 46}+\dfrac {\sin (48-47)}{\sin 47 \sin 48}+...+ \dfrac {\sin (134 -133)}{\sin 133 \sin 134}\right]$

$\Rightarrow \ \dfrac {1}{\sin 1} [\cot 45 -\cot 46+\cot 47-\cot 48+....+\cot 133-\cot 134]$

= we know that

$\cot (\pi -\theta)=-\cot \theta$

using this we get

$\Rightarrow \ \dfrac {1}{\sin 1} [\cot 45 -\cot 46 +\cot 47 -\cot 48..... +\cot 48-\cot 47+\cot 46]$

All terms cancel out except first one

$\Rightarrow \ \dfrac {\cot 45}{\sin 1}=\dfrac {1}{\sin 1}$ .

Let $a_{1},a_{2},..., a_{n}$ be fixed real numbers and define a function $f(x)=(x-a_{1})(x-a_{2})....(x-a_{n})$ .
What is $\displaystyle \lim _{ x\rightarrow a_{1} }f(x)$ ? For some $a\neq a_{1},a_{2},..., a_{n}$ , compute $\displaystyle \lim _{ x\rightarrow a }f(x)$ .

$\begin{array}{l} f\left( x \right) =\left( { X-{ a_{ 1 } } } \right) \left( { X-{ a_{ 2 } } } \right) ..........\left( { X-{ a_{ n } } } \right) \\ { \lim }_{ x\to a }f\left( x \right) =\left( { { a_{ 1 } }-{ a_{ 1 } } } \right) \left( { { a_{ 1 } }-{ a_{ 2 } } } \right) ..........\left( { { a_{ 1 } }-{ a_{ n } } } \right) \\ 0.\left( { { a_{ 1 } }-{ a_{ 2 } } } \right) ......... \\ =0 \\ { \lim }_{ x\to a }f\left( x \right) =\left( { { a_{ 1 } }-{ a_{ 1 } } } \right) \left( { { a_{ 1 } }-{ a_{ 2 } } } \right) ..........\left( { { a_{ 1 } }-{ a_{ n } } } \right) =0 \end{array}$

Hence, this is the answer.

If $\displaystyle\frac{1}{1!10!}+\frac{1}{2!9!}+\frac{1}{3!10!}+...+\frac{1}{1!10!}=\frac{2}{k!}(2^{k-1}-1)$ then find the value of k.

Multiplying and dividing by

$11!$ we get

$\left\{ \dfrac { 1 }{ 11! } \right\} \{ \dfrac { 11! }{ 1!10! } +\dfrac { 11! }{ 2!9! } +\dfrac { 11! }{ 3!8! } +............+\dfrac { 11! }{ 10!1! } \}$

$\Rightarrow\left\{ \dfrac { 1 }{ 11! } \right\} \left\{ { 11 }_{ { C }_{ 1 } }+{ 11 }_{ { C }_{ 2 } }+{ 11 }_{ { C }_{ 3 }.................... }+{ 11 }_{ { C }_{ 10 } } \right\}$

$\Rightarrow \left\{ \dfrac { 1 }{ 11! } \right\} \left\{ { 2 }^{ 11 }-2 \right\}=\left\{ \dfrac { 2 }{ 11! } \right\} \left\{ { 2 }^{ 10 }-1 \right\}$

by comparing we get

$\Rightarrow k=11$

If $\beta \, \neq \, 1$ be any nth root of unity then prove that $1 \, + \, 3\beta \, + \, 5\beta^2 \, + \, ..... \, + \, n \, terms \, = \, -\dfrac{2n}{1 \, - \, \beta}$

$\beta \, = \, (1)^{1/n}\Rightarrow \beta^n \, = \, 1$ ...(1)
If S be the sum of the given series which is arithmetico-geometric series ,then

$S \, = \, 1 \, + \, 3\beta \, + \, 5\beta^2 \, + \,...(2n \, - \, 1)\beta^{n \, - \, 1}$

$\beta S \, = \, \beta \, + \, 3\beta^2 \, + \, ...(2n \, - \, 3)\beta^{n \, - \, 1} \, + \, (2n \, - \, 1)\beta^n$

$S(1 \, - \, \beta) \, = \, 1 \, + \, [2\beta \, + \, \beta^2 \, + \, ...(n \, - \, 1)\, \,terms] \, - \,(2n \, - \,1) \, , \,by(1)$

$= \, 2 \, + \, 2\beta \, + \, 2\beta^2 \, + \, ...n \, \, terms \, \, \, - \, 2n$

$= \, 2\dfrac{(1 \, - \,\beta^n)}{1 \, - \,\beta} \, - \, 2n \, = \, 0 \, - \, 2n\, , \,by \,(1)$

$\therefore \, \, \, S \, = \, \dfrac{- \,2n}{1 \, - \, \beta}$

Let a sequence be defined by $a_1 = 1, a_2 = 1$ and, $a_n=a_{n-1} + a_{n - 2}$ for all $n > 2$ ,
find $\dfrac{a_{n + 1}}{a_n}$ for $n = 1, 2, 3, 4$ .

We have,

$a_1 = 1, a_2 = 1$
and,

$a_n = a_{n - 1} + a_{n -2}$ for all

$n > 2$

Putting

$n = 3, 4$ and

$5$ , we get

$a_3 = a_2 + a_1 = 1 + 1 = 2$

$a_4 = a_3 + a_2 = 2 + 1 = 3$
and,

$a_5 = a_4 + a_3 = 2 + 1 = 5$

Thus, we have

$a_1 = 1, a_2 = 1, a_3 = 2, a_4 = 3$ and

$a_5 = 5$

Now, putting

$n = 1, 2, 3$ and

$4$ in

$\dfrac{a_{n + 1}}{a_n}$ , we get

$\dfrac{a_2}{a_1} = \dfrac{1}{1} = 1$

$[\because a_1 = a_2 = 1]$

$\dfrac{a_3}{a_2} = \dfrac{2}{1} = 2$

$[\because a_2 = 1$ and

$a_3 = 2]$

$\dfrac{a_4}{a_3} = \dfrac{3}{2}$

$[\because a_4 =3$ and

$a_3 = 2]$

$\dfrac{a_5}{a_4} = \dfrac{5}{3}$

$[\because a_4 =3$ and

$a_5 = 5]$

A sequence is defined by $a_n = n^3 - 6n^2 + 11n - 6$ . Show that the first three terms of the sequence are zero and all other terms are positive.

We have,

$a_n = n^3 - 6n^2 + 11n - 6$

Putting

$n = 1, 2, 3$ we get

$a_1 = (1)^3 - 6 \times 1^2 + 11 \times 1 - 6 = 1 - 6 + 11 - 6 = 12 - 12 = 0$

$a_2 = 2^3 - 6 \times 2^2 + 11 \times 2 - 6 = 8 - 24 + 22 - 6 = 30 - 30 = 0$

and,

$a_3 = 3^3 - 6 \times 3^2 + 11 \times 3 - 6 = 27 - 54 + 33 - 6 = 60 - 60 = 0$

Thus, we have

$a_1 = a_2 = a_3 = 0$

We observe that

$a_n = n^3 - 6n^2 + 11n - 6$ is a cubic polynominal in

$n$ and it vanished for

$n = 1, 2$ and

$3$ .

Therefore, by factor theorem

$(n - 1), (n - 2)$ and

$(n - 3)$ are factors of

$a_n$ .

Thus, we have

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

In this expression, if we substitute any value of

$n$ which is greater than

$3$ , then each factor on the

$RHS$ is positive.

Therefore,

$a_n > 0$ for all

$n > 3$ .

Hence, first three of them sequence are zero and all other terms are positive.

$\underset{n \rightarrow \infty}{lim} \dfrac{1^P + 2^P + 3^P + ....+ n^P}{n^{P+ 1}}$ equals-

$\underset{n \rightarrow \infty}{lim} \dfrac{1^P + 2^P + 3^P + .... +n^P}{n^{P+ 1}}$

$=\underset{n \rightarrow \infty}{lim}\dfrac{1}{n} \dfrac{1^P + 2^P + 3^P + .... +n^P}{n^{P}}$

$=\underset{n \rightarrow \infty}{lim}\dfrac{1}{n}\left( \dfrac{1^P}{n^p} + \dfrac{2^P}{n^p} + \dfrac{3^P}{n^P} + .... +\dfrac{n^P}{n^{P}}\right)$

$=\underset{n \rightarrow \infty}{lim}\dfrac{1}{n}\sum\limits_{r=1}^{n}\left(\dfrac{r}{n}\right)^P$

$=\displaystyle\int\limits_{0}^{1}x^Pdx$ [ From the definition of the definite integral]

$=\dfrac{1}{P+1}$ .

Evaluate: $\sum_\limits{r=1}^n (3r-1)(r+1)$

$\sum_{r=1}^{n}(3r-1)(r+1)$

$=\sum_{r=1}^{n}(3r^2+2r-1)$

$=\sum_{r=1}^{n}3r^2+\sum_{r=1}^{n}2r-\sum_{r=1}^{n}1$

$=3\sum_{r=1}^{n}r^2+2\sum_{r=1}^{n}r-1\sum_{r=1}^{n}1$

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

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

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

$=\dfrac{2n^3+n^2+2n^2+n+2n^2}{2}$

$=\dfrac{2n^3+5n^2+n}{2}$

Evaluate $\sum _{ k=1 }^{ 11 }{ \left( 2+{ 3 }^{ k } \right) }$

$\displaystyle \sum_{K=1}^{11}(2+3^{K})$

$= \displaystyle \sum_{K=1}^{11}2+\sum_{K=1}^{11}3^{K}$

$=11\times 2+\dfrac{(3)(3^{11}-1)}{(3-1)}$

$=22+\dfrac{3}{2}(3^{11}-1)$

Solve :
$\dfrac {1}{1.2} + \dfrac {1}{2.3} + ...... + \dfrac {1}{n(n+1)} = ?$

Consider

$n^{th}$ term=

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

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

$\boxed{n^{th}term=\dfrac{1}{n}-\dfrac{1}{(n+1)}}$

$\Rightarrow \dfrac{1}{1.2}=\dfrac{1}{1}-\dfrac{1}{2}$

$\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}$

$\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4}$

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

On adding all terms we gen

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

$\Rightarrow \left [\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....\dfrac{1}{n.(n+1)}\right]=1-\dfrac{1}{n+1}$

$\boxed{\left[\dfrac{1}{1.2}+\dfrac{1}{2.3}+....\dfrac{1}{n.(nh)}\right]=\dfrac{(n+1)-1}{(n+1)}=\left (\dfrac{n}{n+1}\right)}$

Obtain $\sum\limits_{r = 2}^{10} {\left( {4{r^2} - 28r + 4a} \right)}$

$\displaystyle\sum^{10}_{r=2}(4r^2-28r+4a)$

$=4(2^2+3^2+.....+10^2)-28(2+3+...….+10)+(4a+4a+....9 times)$

$=4[(1^2+2^2+3^2+.....+10^2)-1]-28[(1+2+3+.....+10)-1]+36a$

$=4\left[\dfrac{10\times (10+1)\times (2\times 10+1)}{6}-1\right]-28\left(\dfrac{10\times 11}{2}-1\right)+36a$

$=4\times 384-28\times 54+36a$

$=1536-1512+36a$

$=36a+24$ .

Find the sum to $n$ terms of the series
$1.2^{2}+2.3^{2}+3.4^{2}+...$

$1,2^{2}+2,3^{2}+3,4^{2}+$ ….

$Tn \Rightarrow n \left(n+1\right)^{2}$ (General Term of this series)

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

$Tn=n^{3}+2n^{2}+n$

$Sn= \sum Tn= \sum n^{3}+\sum 2n^{2}+\sum n$

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

$\Rightarrow \dfrac{n\left(n+1\right)}{2} \left[\dfrac{n\left(n+1\right)}{2}\dfrac{2\left(2n+1\right)}{3}+1\right]$

$Sn\Rightarrow \dfrac{n\left(n+1\right)}{2} \left[\dfrac{3\left( n\left(n+1\right)\right)+4\left(2n+1\right)+6}{6}\right)$

Find the sum to n terms of the sequence, $8, 88, 888, 8888,....$ .

$8+88+888+8888........... \\ 8\left[ { 1+11+111+1111+..... } \right] \\ \frac { 8 }{ 9 } \left[ { 9+99+999+..... } \right] \\ \frac { 8 }{ 9 } \left[ { \left( { 10-1 } \right) +\left( { 100-1 } \right) +\left( { 1000-1 } \right) +........to\, \, n\, \, term } \right] \\ \frac { 8 }{ 9 } \left[ { \left\{ { 10+100+1000+.......to\, \, n\, \, term } \right\} -n } \right] \\ \frac { 8 }{ 9 } \left[ { \left( { 10+{ { 10 }^{ 2 } }+{ { 10 }^{ 3 } }+.......+to\, \, n\, \, term } \right) -n } \right] \\ \frac { 8 }{ 9 } \left\{ { \frac { { 10\left( { { { 10 }^{ n-1 } } } \right) } }{ { \left( { 10-1 } \right) } } -n } \right\} \, \, =\, \, \frac { 8 }{ 9 } \, \, \left( { \frac { { { { 10 }^{ (n+1) } }-10-9n } }{ 9 } } \right) \, \, =\, \, \frac { 8 }{ { 81 } } \left( { { { 10 }^{ \left( { n+1 } \right) } }-9n-10 } \right) \, \, Ans. \\$

$\cfrac{1}{1.3.5}+\cfrac{1}{3.5.7}+....$ to $n$ terms.

This is

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

$=\sum { \cfrac { 1 }{ 2 } } \times \cfrac { \left( 2r+3 \right) -\left( 2r-1 \right) }{ \left( 2r-1 \right) \left( 2r+1 \right) \left( 2r+1 \right) } =\cfrac { 1 }{ 2 } \times \left[ \sum { \cfrac { 1 }{ \left( 2r-1 \right) \left( 2r+1 \right) } } -\sum { \cfrac { 1 }{ \left( 2r+1 \right) \left( 2r+3 \right) } } \right]$

$=\cfrac { 1 }{ 2 } \times \begin{bmatrix} \cfrac { 1 }{ 1.3 } -\cfrac { 1 }{ 3.5 } \\ \cfrac { 1 }{ 3.5 } -\cfrac { 1 }{ 5.7 } \\ --- \\ --- \\ -\cfrac { 1 }{ \left( 2n+1 \right) \left( 2n+3 \right) } \end{bmatrix}=\cfrac { 1 }{ 2 } \left[ \cfrac { 1 }{ 3 } -\cfrac { 1 }{ \left( 2n+1 \right) \left( 2n+3 \right) } \right]$

Find the sum to $n$ terms of the series, whose $n$ th term is given by ${(2n-1)}^{2}$ .

Given

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

$=4n^2-4n+1$

Sum of n terms is

$S_n=\sum_{n=1}^{n}{a_n}$

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

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

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

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

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

$=\dfrac{n}{3}[4n^2+2n+4n+2-6n-3]$

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

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

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

Find the sum of series upto $n$ terms whose ${ n }^{ th }$ term is ${\left( {2n - 1} \right)^2}$

Consider the problem

Given

$\begin{array}{l} { a_{ n } }={ \left( { 2n-1 } \right) ^{ 2 } } \\ ={ \left( { 2n } \right) ^{ 2 } }+{ \left( 1 \right) ^{ 2 } }-2\left( { 2n } \right) \left( 1 \right) \\ =4{ n^{ 2 } }+1-4n \\ =4{ n^{ 2 } }-4n+1 \end{array}$

Sum of

$n$ term is

$\begin{array}{l} { s_{ n } }=\sum _{ n=1 }^{ n }{ { a_{ n } } } \\ =\sum _{ n=1 }^{ n }{ 4{ n^{ 2 } }-4n+1 } \\ =\sum _{ n=1 }^{ n }{ 4{ n^{ 2 } }-\sum _{ n=1 }^{ n }{ 4n+\sum _{ n=1 }^{ n } 1 } } \\ =4\sum _{ n=1 }^{ n }{ { n^{ 2 } }-4\sum _{ n=1 }^{ n }{ n+\sum _{ n=1 }^{ n } 1 } } \\ =4\left( { \frac { { n\left( { n+1 } \right) \left( { 2n+1 } \right) } }{ 6 } } \right) -4\left( { \frac { { n\left( { n+1 } \right) } }{ 2 } } \right) +n \\ =n\left( { \frac { { n\left( { n+1 } \right) \left( { 2n+1 } \right) } }{ 6 } } \right) -4\left( { \frac { { n\left( { n+1 } \right) } }{ 2 } +1 } \right) \\ =n\left( { \frac { 2 }{ 3 } \left( { n+1 } \right) \left( { 2n+1 } \right) -2\left( { n+1 } \right) +1 } \right) \\ =n\left( { \frac { { 2\left( { n+1 } \right) \left( { 2n+1 } \right) -6\left( { n+1 } \right) +3 } }{ 3 } } \right) \\ =\frac { n }{ 3 } \left[ { \left( { 2n+2 } \right) \left( { 2n+1 } \right) -6n-6+3 } \right] \\ =\frac { n }{ 3 } \left[ { 4{ n^{ 2 } }+2n+4n+2-6n-3 } \right] \\ =\frac { n }{ 3 } \left[ { 4{ n^{ 2 } }+6n-6n-1 } \right] \\ =\frac { n }{ 3 } \left[ { 4{ n^{ 2 } }-1 } \right] \\ =\frac { n }{ 3 } \left[ { { { \left( { 2n } \right) }^{ 2 } }-{ { \left( 1 \right) }^{ 2 } } } \right] \\ =\frac { n }{ 3 } \left[ { \left( { 2n-1 } \right) \left( { 2n+1 } \right) } \right] \end{array}$

Hence, the required sum is

$\frac { n }{ 3 } \left[ { \left( { 2n-1 } \right) \left( { 2n+1 } \right) } \right]$

Find the sum of $1 + 5 + {5^2} + ....$ upto 8 terms.

$\begin{array}{l} sum\, \, of1+5+{ 5^{ 2 } }+{ 5^{ 3 } }\, \, ...\, \, up\, \, 8th\, \, terms \\ \left( a \right) First\, \, term=1 \\ common\, \, ratio\left( r \right) =5 \\ Sum\, \, of\, \, 8th\, \, term=\frac { { a\left( { { r^{ n } }-1 } \right) } }{ { r-1 } } =\frac { { 1\left( { { 5^{ 8 } }-1 } \right) } }{ { 5-1 } } =\frac { { { 5^{ 8 } }-1 } }{ 4 } \end{array}$

Find the sum of the series $2 ^ { 2 } + 4 ^ { 2 } + 6 ^ { 2 } + \ldots + ( 2 n ) ^ { 2 }$

$2^{2}+4^{2}+6^{2}+...(2n)^{2}$

let

$Tn=(2n)^{2}=4n^{2}$

$\sum Tn = \sum 4n^{2}$

$= 4\sum n^{2}$

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

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

1141738_1246910_ans_bc5b3bafb08f44349cd60148d6ee1b28.jpg

Show that $1^{2}+(1^{2}+2^{2})+(1^{2}+2^{2}+3^{2})+........$ upto $n$ terms $=\dfrac{n(n+1)^{2}(n+2)}{12}, \forall n \in N$

Formuale being used:

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

$\sum_{i=1}^{n}i^2=\dfrac{n(n+1)(2n+1)}{6}$

$\sum_{i=1}^{n}i^3=\dfrac{n^2(n+1)^2}{4}$

Given,

$1^2+(1^2+2^2)+(1^2+2^2+3^2)+..............+(1^2+2^2+3^2+........+n^2)$

$=\sum_{i=1}^{n}(1^2+2^2+3^2+........+i^2)$

$=\sum_{i=1}^{n}\sum_{j=1}^{i}j^2$

$=\sum_{i=1}^{n}\dfrac{i(i+1)(2i+1)}{6}$

$=\sum_{i=1}^{n}\dfrac{2i^3+3i^2+i}{6}$

$=\sum_{i=1}^{n}\dfrac{1}{3}i^3+\sum_{j=1}^{n}\dfrac{1}{2}j^2+\sum_{k=1}^{n}\dfrac{1}{6}k$

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

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

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

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

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

Find the sum of the series whose $n^{th}$ term is :
$2n^{3}+3n^{2}-1$

$n^{th}$ term is

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

$S_{n}=\sum_{k=1}^{n}(2k^{3}+3k^{2}-1)$

$=\sum_{k=1}^{n}2k^{3}+\sum_{k=1}^{n}3k^{2}-\sum_{k=1}^{n}1$

$=2\sum_{k=1}^{n}k^{3}+3\sum_{k=1}^{n}k^{2}-\sum_{k=1}^{n}1$

$=2(\frac{n^{2}(n+1)^{2}}{4})+3(\frac{n(n+1)(2n+1)}{6})-n$

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

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

$=\frac{n^{4}+n^{2}+2n^{3}+2n^{3}+3n^{2}+n-2n}{2}=\frac{n^{4}+4n^{3}+4n^{2}-n}{2}$

1186827_1292317_ans_014a1721b30d41aabbc00b6bef156d31.jpg

$\left( r+1 \right) ^{ 2 }+\left( r+2 \right) ^{ 2 }+....+{ n }^{ 2 }$

We have,

${{\left( r+1 \right)}^{2}}+{{\left( r+2 \right)}^{2}}+......+{{n}^{2}}$

Let $r+1=m$

So,

${{m}^{2}}+{{\left( m+1 \right)}^{2}}+{{\left( m+2 \right)}^{2}}+......+{{n}^{2}}$

$\Rightarrow \sum\limits_{1}^{n}{{{m}^{2}}}$

$\Rightarrow \sum\limits_{1}^{n}{{{\left( r+1 \right)}^{2}}}$

Hence, this is the answer.

Find sum of $n$ terms of following series ?
$1 + 3 x + 5 x ^ { 2 } + 7 x ^ { 3 } + \dots$

$S=1+3x+{ 5x }^{ 2 }+{ 7x }^{ 3 }+......+\left( 2n+1 \right) { x }^{ n }$

$\underline { xS=x+{ 3x }^{ 2 }+{ 5x }^{ 3 }+......+\left( 2n-1 \right) { x }^{ n }+{ \left( 2n+1 \right) }^{ n+1 }x }$

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

$=\left[ 1+\left( 2n+1 \right) { x }^{ n+1 } \right] +2x\left( 1+x+{ x }^{ 2 }+......{ x }^{ n } \right)$

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

For all real numbers $a, b$ and positive integer $n$ prove that:
$(a+b)^{n}=^{n}C_{0}a^{n}+^{n}C_{1}a^{n-1}b+^{n}C_{1}a^{n-2}b^{2}+..........+^{n}C_{n}b^{n}$ .

$\begin{array}{l} { \left( { a+b } \right) ^{ n } }{ =^{ n } }{ c_{ 0 } }{ a^{ n } }{ =^{ n } }{ c_{ 1 } }{ a^{ n-1 } }b{ +^{ n } }{ c_{ 2 } }{ a^{ n-2 } }{ b^{ 2 } }+.......{ +^{ n } }{ c_{ n } }{ b^{ n } } \\ \Rightarrow { \left( { a+b } \right) ^{ 1 } }{ =^{ 1 } }{ c_{ 0 } }{ a^{ 1 } }{ +^{ 1 } }{ c_{ 1 } }{ a^{ 0 } }b=\left( { a+b } \right) \\ \Rightarrow \left( { a+{ b^{ 2 } } } \right) { =^{ 2 } }{ c_{ 0 } }{ a^{ 2 } }{ +^{ 2 } }{ c_{ 1 } }ab{ +^{ 2 } }{ c_{ 2 } }{ b^{ 2 } }={ a^{ 2 } }+2ab+{ b^{ 2 } } \\ \Rightarrow { \left( { a+b } \right) ^{ 3 } }=3{ c_{ 0 } }{ a^{ 3 } }{ +^{ 3 } }{ c_{ 1 } }{ a^{ 2 } }b{ +^{ 3 } }{ c_{ 2 } }a{ b^{ 2 } }+3{ c_{ 3 } }{ a^{ 0 } }{ b^{ 3 } }={ a^{ 3 } }+3{ a^{ 2 } }b+3a{ b^{ 2 } }+{ b^{ 3 } } \\ \Rightarrow \therefore { \left( { a+b } \right) ^{ n } }{ =^{ n } }{ c_{ 0 } }{ a^{ n } }{ =^{ n } }{ c_{ 1 } }{ a^{ \left( { n-1 } \right) } }b{ +^{ n } }{ c_{ 2 } }{ a^{ \left( { n-2 } \right) } }{ b^{ 2 } }+......{ +^{ n } }{ c_{ n } }{ b^{ n } }. \end{array}$

Prove that
$\dfrac {C_{1}}{C_{0}}+\dfrac {2C_{2}}{C_{1}}+\dfrac {3C_{3}}{C_{2}}+....+\dfrac {n.C_{n}}{C_{n-1}}=\dfrac {n(n+1)}{2}$

We have,

$\dfrac{kC_k}{C_{k-1}}=\dfrac{k \binom {n}{k}} {\binom{n}{k-1}}$

$=k\cdot \dfrac{n!}{k!(n-k)!}\dfrac{(k-1)!(n-k+1)!}{n!}$

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

$=n-k+1$

$k=1\rightarrow \dfrac{C_1}{C_0}=n$

$k=2\rightarrow \dfrac{2C_2}{C_1}=n-1$

$k=3\rightarrow \dfrac{3C_3}{C_2}=n-2$

$k=n\rightarrow \dfrac{nC_n}{C_{n-1}}=1$

On adding all the above terms, we get,

$\dfrac{C_1}{C_0}+\dfrac{2C_2}{C_1}+\dfrac{3C_3}{C_2}+.........+\dfrac{nC_n}{C_{n-1}}=n+(n-1)+(n-2)+......+1$

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

Hence proved.

Write the first five terms of the sequence and obtain the corresponding series:
$a_{1}=a_{2}=2,a_{n}=a_{n-1}-1,n > 2$

Given,

$a_1=a_2=2$

Formula,

$a_n=a_{n-1}-1$

$a_3=a_{3-1}-1$

$=a_2-1=2-1=1$

$a_4=a_{4-1}-1$

$=a_3-1=1-1=0$

$a_5=a_{5-1}-1$

$=a_4-1=0-1=-1$

Hence the first five terms of the series are:

$2,2,1,0,-1$

Evaluate:
$\dfrac{1+\dfrac{1}{2!}+\dfrac{1}{3!}+.......}{1+\dfrac{1}{3!}+\dfrac{1}{5!}+.......}$ .

We've,

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

And,

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

Then

$e+e^{-1}=2(1+\dfrac{1}{2!}+\dfrac{1}{4!}+......)$ .....(1) and

$e-e^{-1}=2(1+\dfrac{1}{3!}+\dfrac{1}{5!}+......)$ .....(2).

Now,

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

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

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

Find the coefficient of $x$ in the expansion of $\left( 1+\dfrac { x }{ 1! } +\dfrac { x^2 }{ 2! } +\dfrac { x^3 }{ 3! } +...+\dfrac { x^n }{ n! } \right)$

$1 + \cfrac{x}{1!} + \cfrac{{x}^{2}}{2!} + \cfrac{{x}^{3}}{3!} + ..... + \cfrac{{x}^{n}}{n!}$

Therefore,

Coefficient of

$x = \cfrac{1}{1!} = \cfrac{1}{1} = 1$

Hence the coefficient of

$x$ in the given expression is

$1$ .

The sum, $\sum _{ n=1 }^{ 7 }{ \cfrac { n\left( n+1 \right) \left( 2n+1 \right) }{ 4 } }$ is equal to ________

$= \sum_{n = 1}^7 \dfrac{n (n + 1) (2n + 1)}{4}$

$=\dfrac{1}{4} \sum_{n = 1}^7 (2n^3 + 3n^2 + n)$

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

$=\dfrac{1}{4} \left[2 . \left(\dfrac{7. (7 + 1)}{2}\right)^2 + 3 \dfrac{7 (7 + 1)(2.7 + 1)}{6} + \dfrac{7(7 + 1)}{2} \right]$

$=\dfrac{1}{4} \left[2 \times \dfrac{7^2 . 8^2}{4} + \dfrac{7 \times 8 \times 15}{2} + \dfrac{7 \times 8}{2}\right)$

$=\dfrac{1}{4} (1568 + 420 + 28)$

$= 504$

Determine whether the following series is convergent or divergent
$\displaystyle \frac {1}{1^p}+\frac {1}{3^p}+\frac {1}{5^p}+\frac {1}{7^p}+.....$

We have to find the given series is convergent or not

$\dfrac{1}{1^p}+\dfrac{1}{3^p}+\dfrac{1}{5^p}+\dfrac{1}{7^p}+....+\dfrac{1}{(2n-1)^p}$

$u_n=\dfrac{1}{(2n-1)^p};$

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

$\Rightarrow$

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

$\displaystyle\lim_{n\rightarrow\infty}\dfrac{u_{n+1}}{u_n}=\dfrac{1-0}{1+0}=1$

$\therefore$ Ratio test fails.

Now, take

$v_n=\dfrac{1}{n^p}$

$\dfrac{u_n}{v_n}=\dfrac{n^p}{(2n-1)^p}=\dfrac{1}{2^p\left(1-\dfrac{1}{2n}\right)^p}$

$\Rightarrow$

$\displaystyle\lim_{n\rightarrow\infty}\dfrac{u_n}{v_n}=\dfrac{1}{2^p}$

Which is non-zero and finite.

$\therefore$ By comparison test,

$\sum u_n$ and

$\sum v_n$ both converge or both diverge.

But by

$p-$ series test,

$\sum v_n=\sum\dfrac{1}{n^p}$ converges, when

$p>1$ and diverges when

$p\le 1$

$\therefore$

$\sum u_n$ is convergent if

$p>1$ and divergent if

$p\le 1.$

Find the general term and the sum of $n$ terms of the series:
$2,5,12,31,86,......$

1979060_1744873_ans_0d3fdf033bb245dea90372b4db15e5d4.jpg

Determine whether the following series is convergent or divergent
$\displaystyle 1+ \frac {1}{2^2}+\frac {2^2}{3^3}+\frac {3^3}{4^4}+\frac {4^4}{5^5}....$

Neglecting the first term, the series is

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

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

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

Take

$v_n=\dfrac{1}{n}$

$\therefore$

$\displaystyle\lim_{n\rightarrow\infty}\dfrac{u_n}{v_n}=\lim_{n\rightarrow \infty}\dfrac{1}{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{1}{n}\right)^n}=\lim_{n\rightarrow \infty}\dfrac{1}{\left(1+\dfrac{1}{n}\right)^n.1}=\dfrac{1}{e}$

Which is finite and

$\sum v_n=\sum\dfrac{1}{n}$ is divergent by

$p-$ series test

$(p=1)$

$\therefore$

$\sum u_n$ is divergent.

Determine whether the following series is convergent or divergent
$\displaystyle1 + \frac {2}{5}x+\frac {6}{9}x^2+\frac {14}{17}x^3+....+\frac {2^n-2}{2^n+1}x^{n-1}+....$

Neglecting the first term,

$u_n=\left(\dfrac{2^n-2}{2^n+1}\right)x^{n-1},\,(n\ge 2)$ and

$u_n$ are all positive.

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

$\displaystyle\lim_{n\rightarrow\infty}\dfrac{u_{n+1}}{u_n}=\lim_{n\rightarrow \infty}\left(\dfrac{2^{n+1}-2}{2^{n+1}+1}\right)\times\left(\dfrac{2^n+1}{2^n-2}\right)x$

$=\displaystyle\lim_{n\rightarrow \infty}\left[\dfrac{2^{n+1}(1-1/2^n)}{2^{n+1}(1+1/2^{n+1})}\times \dfrac{2^(1+1/2^n)}{2^n(1-2/2^n)}.x\right]$

$=x$

Hence, by ratio test,

$\sum u_n$ converges if

$x<1$ and diverges if

$x>1$ .

When

$x=1,$ the test fails.

Then,

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

$\Rightarrow$

$\displaystyle\lim_{n\rightarrow\infty}u_n=1\ne 0$

$\therefore$

$\sum u_n$ diverges.

Hence,

$\sum u_n$ is convergent when

$x<1$ and divergent

$x>1$

If $y=x+ x^2 + x^3 + ...\infty$ , where $|x| < 1$ , then prove that

$x= \left( \dfrac{y}{1+y} \right)$

$y= x(1+x+x^2 + ...\infty)$

$\implies y(1-x) = x$

$x= \left( \dfrac{y}{1+y} \right)$

Prove that $\left(\displaystyle a+\frac{1}{b+}\frac{1}{a+}\frac{1}{b+}\frac{1}{a+}....\right)\left(\displaystyle\frac{1}{b+}\frac{1}{a+}\frac{1}{b+}\frac{1}{a+}....\right)=\displaystyle \frac{a}{b}$ .

Let

$x=a+\cfrac { 1 }{ b+ } \cfrac { 1 }{ a+ } \cfrac { 1 }{ b+ } \cfrac { 1 }{ a+ } .....$

$=a+\cfrac { 1 }{ b+\cfrac { 1 }{ a+x-a } } =a+\cfrac { x }{ bx+1 }$

Thus

$b{ x }^{ 2 }+x=abx+x+a$

$\Rightarrow b{ x }^{ 2 }-abx-a=0$

$\Rightarrow x=\cfrac { ab+\sqrt { { a }^{ 2 }{ b }^{ 2 }+4ab } }{ 2b }$

$\Rightarrow y=\cfrac { 1 }{ b+\cfrac { 1 }{ a+y } } =\cfrac { a+y }{ ab+by+1 }$

$\Rightarrow aby+b{ y }^{ 2 }+y=a+y$

$\Rightarrow b{ y }^{ 2 }+aby-a=0$

$\Rightarrow y=\cfrac { -ab+\sqrt { { a }^{ 2 }{ b }^{ 2 }+4ab } }{ 2b }$

$\Rightarrow xy=\cfrac { { a }^{ 2 }{ b }^{ 2 }+4ab-{ a }^{ 2 }{ b }^{ 2 } }{ 4{ b }^{ 2 } }$

$=\cfrac { a }{ b }$

Hence, proved.

Find the sum of the infinite series $1 + \dfrac{2}{3}.\dfrac{1}{2} + \dfrac{2.5}{3.6}(\dfrac{1}{2})^{2} + \dfrac{2.5.8}{3.6.9}(\dfrac{1}{2})^{3} + ....\infty$

The given infinite series is,

$S=1+\frac { 2 }{ 3 } .\frac { 1 }{ 2 } +\frac { 2 }{ 3 } .\frac { 5 }{ 6 } .{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }+\frac { 2 }{ 3 } .\frac { 5 }{ 6 } .\frac { 8 }{ 9 } .{ \left( \frac { 1 }{ 2 } \right) }^{ 3 }------------$ (1)

Now, expansion of binomial series

${ \left( 1+x \right) }^{ n }$ is given by,

${ \left( 1+x \right) }^{ n }=1+nx+\frac { n\left( n-1 \right) }{ 2! } { x }^{ 2 }+\frac { n\left( n-1 \right) \left( n-2 \right) }{ 3! } { x }^{ 2 }+----------$ (2)

Comparing equation (1) and (2), first term of both equations is same.

Thus, comparing second term of both equations, we get,

$nx=\frac { 2 }{ 3 } \times \frac { 1 }{ 2 }$

$\therefore nx=\frac { 1 }{ 3 }$

Squaring both sides, we get,

$\therefore { n }^{ 2 }{ x }^{ 2 }=\frac { 1 }{ 9 }$

$\therefore { x }^{ 2 }=\frac { 1 }{ 9{ n }^{ 2 } }$ (3)

Comparing third term of both series, we get,

$\frac { n\left( n-1 \right) }{ 2! } { x }^{ 2 }=\frac { 2 }{ 3 } .\frac { 5 }{ 6 } .{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }$

$\therefore \frac { n\left( n-1 \right) }{ 2 } { x }^{ 2 }=\frac { 5 }{ 9 } .\frac { 1 }{ 4 }$

$\therefore \frac { n\left( n-1 \right) }{ 2 } { x }^{ 2 }=\frac { 5 }{ 36 }$

From equation (3),

$\therefore \frac { n\left( n-1 \right) }{ 2 } .\frac { 1 }{ 9{ n }^{ 2 } } =\frac { 5 }{ 36 }$

$\therefore n\left( n-1 \right) \times \frac { 1 }{ { n }^{ 2 } } =\frac { 5\times 18 }{ 36 }$

$\therefore \frac { \left( n-1 \right) }{ { n } } =\frac { 5 }{ 2 }$

$\therefore \left( n-1 \right) =\frac { 5 }{ 2 } n$

$\therefore \frac { 5 }{ 2 } n-n=-1$

$\therefore \frac { 3 }{ 2 } n=-1$

$\therefore n=\frac { -2 }{ 3 }$

From equation (1),

$\frac { -2 }{ 3 } x=\frac { 1 }{ 3 }$

$\therefore x=\frac { -1 }{ 2 }$

Thus, sum of series is given by,

$S={ \left( 1+x \right) }^{ n }$

$S={ \left( 1-\frac { 1 }{ 2 } \right) }^{ \frac { -2 }{ 3 } }$

$S={ \left( \frac { 1 }{ 2 } \right) }^{ \frac { -2 }{ 3 } }$

$\therefore S=\frac { 1 }{ { \left( 2 \right) }^{ \frac { -2 }{ 3 } } }$

$\therefore S={ \left( 2 \right) }^{ \frac { 2 }{ 3 } }$

$\therefore S={ \left( 4 \right) }^{ \frac { 1 }{ 3 } }$

Find sum of $3+33+333+3333+$ ....... up to n terms.

From the given series, take common factor 3

Thus

$S=3(1+11+111+1111+...............nth term)$

$S=3\left\{ 1+\left( 1+10 \right) +\left( 1+10+100 \right) +\left( 1+10+100+1000 \right) +...to\quad n\quad terms \right\}$

Thus the nth term.

${ t }_{ n }=1+10+100+..............+{ 10 }^{ n-1 }$

This is a G.P. where first term is 1 and common ratio is 10 ,

Thus sum of n terms is

${ t }_{ n }=\dfrac { 1\left( 1-{ 10 }^{ n } \right) }{ 1-10 } \\ { t }_{ n }=\dfrac { \left( { 10 }^{ n }-1 \right) }{ 9 }$

Thus

$S=3\sum _{ i=1 }^{ n }{ { t }_{ n } } =3\sum _{ i=1 }^{ n }{ \dfrac { { 10 }^{ n }-1 }{ 9 } }$

Hence the sum,

$S=3\sum { \dfrac { { 10 }^{ n } }{ 9 } - } 3\sum { \dfrac { 1 }{ 9 } }$

$S=3\left\{ \dfrac { 10\left( 1-{ 10 }^{ n } \right) }{ \left( 1-10 \right) .9 } -\left( \dfrac { 1 }{ 9 } \right) .n \right\}$

$S=3\left\{ 10.\dfrac { \left( { 10 }^{ n }-1 \right) }{ 81 } -\dfrac { n }{ 9 } \right\}$

Thus,

$S=\left\{ 10.\dfrac { \left( { 10 }^{ n }-1 \right) }{ 27 } -\dfrac { n }{ 3 } \right\} \\ =\dfrac { { 10 }^{ n+1 }-10-9n }{ 27 }$

Find the sum of the series $(1^2+1)1!+(2^2+1)2!+(3^2+1)3!+......(n^2+1)n!$

$n!=1\times2\dots n$

$(n+1)n! = 1\times2\dots n\times(n+1)=(n+1)!$

$(P)$
------------------------------------------------------------------------

$S=\displaystyle\sum_{k=1}^{n}(k^2+1)k!$

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

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

$=\displaystyle\sum_{k=1}^{n}(k+1)^2k!-2k\cdot k!$

$=\displaystyle\sum_{k=1}^{n}(k+1)\cdot(k+1)!-2k\cdot k!$ (using

$(P)$ )

$=\displaystyle\sum_{k=1}^{n}[(k+1)\cdot(k+1)!-k\cdot k!] - \displaystyle\sum_{k=1}^{n}k\cdot k!$

$=(n+1)(n+1)!-1\cdot1! -\displaystyle\sum_{k=1}^{n}(k+1-1)k!$

$=(n+1)(n+1)!-1 -\left[\displaystyle\sum_{k=1}^{n}(k+1)k!-k!\right]$

$=(n+1)(n+1)!-1 -\left[\displaystyle\sum_{k=1}^{n}(k+1)!-k!\right]$ (using

$(P)$ )

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

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

$=\boxed{n(n+1)!}$

$3+33+333+.....+$ n terms.

$\begin{array}{l} S=3+33+333+.....+n\, \, term\, \, is \\ S=3\left( { 1+11+111+.....+n } \right) \\ S=\frac { { 3\times 9 } }{ 9 } \left( { 1+11+111+.......n } \right) \\ S=\frac { 3 }{ 9 } \left( { 9+99+999+.....+n } \right) \\ S=\frac { 3 }{ 9 } \left( { 10-1 } \right) +\left( { 100-1 } \right) +\left( { { { 10 }^{ 2 } }-1 } \right) +...+n \\ S=\frac { 3 }{ 9 } \left[ { \left( { 10+{ { 10 }^{ 2 } }+{ { 10 }^{ 3 } }+{ { .....10 }^{ n } } } \right) -\left( n \right) } \right] \\ The\, \, sense\, \, in\, GP,\, \, then \\ S=\frac { 1 }{ 3 } \left( { \frac { { 10\left( { { { 10 }^{ n } }-1 } \right) } }{ { 10-1 } } -n } \right) \\ S=\frac { 1 }{ 3 } \left( { \frac { { { { 10 }^{ n+1 } }-10 } }{ 9 } -n } \right) \\ S=\frac { 1 }{ 3 } \left( { \frac { { { { 10 }^{ n+1 } }10-9n } }{ 9 } } \right) \\ S=\frac { 1 }{ { 27 } } \left[ { { { 10 }^{ n+1 } }-9n-10 } \right] \end{array}$

The sum of the series $\displaystyle \sum _{ k=1 }^{ 62 }{ \frac { 1 }{ \left( k+2 \right) \sqrt { k+1 } +\left( k+1 \right) \sqrt { k+2 } } }$ .

$\sum_{k =1}^{62} \dfrac{1}{(k+2) \sqrt{k+1}+ (k+1) \sqrt{k+2}}$

$\sum_{k=1}^{62} \dfrac{(k+2)-(k+1)}{(k+2) \sqrt{k+1}+ (k+1) \sqrt{k+2}}$

$S_{n} = \sum_{k=1}^{62} \left[ \dfrac{1}{\sqrt{k+1}}- \dfrac{1}{\sqrt{k+2}} \right]$

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

$S_{2}= \dfrac{1}{\sqrt{3}}- \dfrac{1}{\sqrt{4}}$

$S_{3}= \dfrac{1}{\sqrt{4}}- \dfrac{1}{\sqrt{5}}$

$S_{62}=\dfrac{1}{\sqrt{63}}- \dfrac{1}{\sqrt{64}}$

Adding the above sums, we get

$S_{62} = \sum_{k=1}^{62} = \dfrac{1}{\sqrt{2}}- \dfrac{1}{\sqrt{64}} = \dfrac{1}{\sqrt{2}}- \dfrac{1}{8}= \dfrac{8- \sqrt{2}}{8 \sqrt{2}}= \dfrac{8 \sqrt{2}-2}{8.2}$

$=\dfrac{2 (4 \sqrt{2}-1)}{8 \times 2} = \dfrac{4 \sqrt{2}-1}{8} = S_{62}$

How many three-digit numbers are there with no digit repeated?

$\begin{array}{l} Digits\, \, \, we\, \, have \\ 0,1,2,3,4,5,6,7,8,9 \\ 9=1-9 \\ 9=0-9 \\ 8= \\ (i)\, \, for\, I\, we\, \, have\, \, 9\, \, choices\, \, 1-9 \\ (ii)\, \, for\, \, II\, \, we\, \, have\, \, \, 9\, \, choices\, \, except\, I \\ (iii)\, \, for\, \, III\, \, we\, \, have\, \, 8\, \, choices \end{array}$

Find the sum of the series $1.3^{2}+2.5^{2}+3.7^{2}+$ to $n$ terms.

$\begin{array}{l} S={ 1.3^{ 2 } }+{ 2.5^{ 2 } }+{ 3.7^{ 2 } }+...... \\ S=\sum _{ r=1 }^{ n }{ r{ { \left( { 2r+1 } \right) }^{ 2 } } } \\ =\sum { r\left( { 4{ r^{ 2 } }+1+4r } \right) } =4\sum { { r^{ 3 } } } +\sum r+4\sum { { r^{ 2 } } } \\ =\frac { { 4{ { \left[ { n\left( { n+1 } \right) } \right] }^{ 2 } } } }{ 4 } +\frac { { n\left( { n+1 } \right) } }{ 2 } +4\frac { { n\left( { n+1 } \right) \left( { 2n+1 } \right) } }{ 6 } \\ ={ n^{ 2 } }{ \left( { n+1 } \right) ^{ 2 } }+\frac { { n\left( { n+1 } \right) } }{ 2 } +\frac { 2 }{ 3 } \left( n \right) \left( { n+1 } \right) \left( { 2n+1 } \right) \\ =n\left( { n+1 } \right) \left[ { n\left( { n+1 } \right) +\frac { 1 }{ 2 } +\frac { 2 }{ 3 } \left( { 2n+1 } \right) } \right] \\ =n\left( { n+1 } \right) \left[ { { n^{ 2 } }+\frac { { 7n } }{ 3 } +\frac { 7 }{ 6 } } \right] \\ \end{array}$

Solve:
$\frac{1}{{2.5}} + \frac{1}{{5.8}} + \frac{1}{{8.11}} + ...\frac{1}{{(3n - 1)(3n + 2)}}$

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Sequences And Series - Class 11 Commerce Maths - Extra Questions

Find the following sum.n∑r=1(6r2−2r+6)\sum\limits_{r = 1}^n {\left( {6{r^2} - 2r + 6} \right)}

1.2+2.3x+3.4x2+....∞(|x|<1)1.2 + 2.3x + 3.4x^{2} + ....\infty (|x| < 1).

Write the general term for the given series 1(n−1),13(n−3),15(n−5),....\frac{1}{(n-1)},\frac{1}{3(n-3)},\frac{1}{5(n-5)},....

Solve:n∑r=1(6r2−2r+6)\displaystyle\sum\limits_{r = 1}^n {\left( {6{r^2} - 2r + 6} \right)}

The product of two consecutive positive numbers is 182182. Find the numbers.

Evaluate 1+i2+i4+i6+...+i2n1+i^{2}+i^{4}+i^{6}+...+i^{2n}.

What is the sum of first 20 odd natural numbers?

Sum the following series1+34+716+1564+31256+....1 + \dfrac {3}{4} + \dfrac {7}{16} + \dfrac {15}{64} + \dfrac {31}{256} + .... to infinity.

Find value of the sum 1sin45osin46o+1sin47osin48o+1sin49osin50o+.......+1sin133osin134o\displaystyle \frac{1}{sin45^o sin 46^o} + \frac{1}{sin47^o sin 48^o} + \frac{1}{sin49^o sin 50^o} + ....... + \frac{1}{sin133^o sin 134^o}

If 11!10!+12!9!+13!10!+...+11!10!=2k!(2k−1−1)\displaystyle\frac{1}{1!10!}+\frac{1}{2!9!}+\frac{1}{3!10!}+...+\frac{1}{1!10!}=\frac{2}{k!}(2^{k-1}-1) then find the value of k.

If β≠1\beta \, \neq \, 1 be any nth root of unity then prove that 1+3β+5β2+.....+nterms=−2n1−β1 \, + \, 3\beta \, + \, 5\beta^2 \, + \, ..... \, + \, n \, terms \, = \, -\dfrac{2n}{1 \, - \, \beta}

Let a sequence be defined by a1=1,a2=1a_1 = 1, a_2 = 1 and, an=an−1+an−2a_n=a_{n-1} + a_{n - 2} for all n>2n > 2,find an+1an\dfrac{a_{n + 1}}{a_n} for n=1,2,3,4n = 1, 2, 3, 4.

A sequence is defined by an=n3−6n2+11n−6a_n = n^3 - 6n^2 + 11n - 6. Show that the first three terms of the sequence are zero and all other terms are positive.

limn→∞1P+2P+3P+....+nPnP+1\underset{n \rightarrow \infty}{lim} \dfrac{1^P + 2^P + 3^P + ....+ n^P}{n^{P+ 1}} equals-

Evaluate: n∑r=1(3r−1)(r+1)\sum_\limits{r=1}^n (3r-1)(r+1)

Evaluate 11∑k=1(2+3k)\sum _{ k=1 }^{ 11 }{ \left( 2+{ 3 }^{ k } \right) }

Solve :11.2+12.3+......+1n(n+1)=? \dfrac {1}{1.2} + \dfrac {1}{2.3} + ...... + \dfrac {1}{n(n+1)} = ?

Obtain 10∑r=2(4r2−28r+4a)\sum\limits_{r = 2}^{10} {\left( {4{r^2} - 28r + 4a} \right)}

Find the sum to nn terms of the series1.22+2.32+3.42+...1.2^{2}+2.3^{2}+3.4^{2}+...

Find the sum to n terms of the sequence, 8,88,888,8888,....8, 88, 888, 8888,.....

11.3.5+13.5.7+....\cfrac{1}{1.3.5}+\cfrac{1}{3.5.7}+.... to nn terms.

Find the sum to nn terms of the series, whose nnth term is given by (2n−1)2{(2n-1)}^{2}.

Find the sum of series upto nn terms whose nth{ n }^{ th } term is (2n−1)2{\left( {2n - 1} \right)^2}

Find the sum of 1+5+52+....1 + 5 + {5^2} + .... upto 8 terms.

Find the sum of the series 22+42+62+…+(2n)22 ^ { 2 } + 4 ^ { 2 } + 6 ^ { 2 } + \ldots + ( 2 n ) ^ { 2 }

Show that 12+(12+22)+(12+22+32)+........1^{2}+(1^{2}+2^{2})+(1^{2}+2^{2}+3^{2})+........ upto nn terms =n(n+1)2(n+2)12,∀n∈N=\dfrac{n(n+1)^{2}(n+2)}{12}, \forall n \in N

Find the sum of the series whose nthn^{th} term is :2n3+3n2−12n^{3}+3n^{2}-1

(r+1)2+(r+2)2+....+n2\left( r+1 \right) ^{ 2 }+\left( r+2 \right) ^{ 2 }+....+{ n }^{ 2 }

Find sum of nn terms of following series ?1+3x+5x2+7x3+…1 + 3 x + 5 x ^ { 2 } + 7 x ^ { 3 } + \dots

For all real numbers a,ba, b and positive integer nn prove that:(a+b)n=nC0an+nC1an−1b+nC1an−2b2+..........+nCnbn(a+b)^{n}=^{n}C_{0}a^{n}+^{n}C_{1}a^{n-1}b+^{n}C_{1}a^{n-2}b^{2}+..........+^{n}C_{n}b^{n}.

Prove that C1C0+2C2C1+3C3C2+....+n.CnCn−1=n(n+1)2\dfrac {C_{1}}{C_{0}}+\dfrac {2C_{2}}{C_{1}}+\dfrac {3C_{3}}{C_{2}}+....+\dfrac {n.C_{n}}{C_{n-1}}=\dfrac {n(n+1)}{2}

Write the first five terms of the sequence and obtain the corresponding series:a_{1}=a_{2}=2,a_{n}=a_{n-1}-1,n > 2a_{1}=a_{2}=2,a_{n}=a_{n-1}-1,n > 2

Evaluate:\dfrac{1+\dfrac{1}{2!}+\dfrac{1}{3!}+.......}{1+\dfrac{1}{3!}+\dfrac{1}{5!}+.......}\dfrac{1+\dfrac{1}{2!}+\dfrac{1}{3!}+.......}{1+\dfrac{1}{3!}+\dfrac{1}{5!}+.......}.

Find the coefficient of xx in the expansion of \left( 1+\dfrac { x }{ 1! } +\dfrac { x^2 }{ 2! } +\dfrac { x^3 }{ 3! } +...+\dfrac { x^n }{ n! } \right) \left( 1+\dfrac { x }{ 1! } +\dfrac { x^2 }{ 2! } +\dfrac { x^3 }{ 3! } +...+\dfrac { x^n }{ n! } \right)

The sum, \sum _{ n=1 }^{ 7 }{ \cfrac { n\left( n+1 \right) \left( 2n+1 \right) }{ 4 } } \sum _{ n=1 }^{ 7 }{ \cfrac { n\left( n+1 \right) \left( 2n+1 \right) }{ 4 } } is equal to ________

Determine whether the following series is convergent or divergent\displaystyle \frac {1}{1^p}+\frac {1}{3^p}+\frac {1}{5^p}+\frac {1}{7^p}+.....\displaystyle \frac {1}{1^p}+\frac {1}{3^p}+\frac {1}{5^p}+\frac {1}{7^p}+.....

Find the general term and the sum of nn terms of the series:2,5,12,31,86,......2,5,12,31,86,......

Determine whether the following series is convergent or divergent\displaystyle 1+ \frac {1}{2^2}+\frac {2^2}{3^3}+\frac {3^3}{4^4}+\frac {4^4}{5^5}....\displaystyle 1+ \frac {1}{2^2}+\frac {2^2}{3^3}+\frac {3^3}{4^4}+\frac {4^4}{5^5}....

Determine whether the following series is convergent or divergent\displaystyle1 + \frac {2}{5}x+\frac {6}{9}x^2+\frac {14}{17}x^3+....+\frac {2^n-2}{2^n+1}x^{n-1}+....\displaystyle1 + \frac {2}{5}x+\frac {6}{9}x^2+\frac {14}{17}x^3+....+\frac {2^n-2}{2^n+1}x^{n-1}+....

If y=x+ x^2 + x^3 + ...\inftyy=x+ x^2 + x^3 + ...\infty, where |x| < 1|x| < 1, then prove thatx= \left( \dfrac{y}{1+y} \right)x= \left( \dfrac{y}{1+y} \right)

Find the sum of the infinite series 1 + \dfrac{2}{3}.\dfrac{1}{2} + \dfrac{2.5}{3.6}(\dfrac{1}{2})^{2} + \dfrac{2.5.8}{3.6.9}(\dfrac{1}{2})^{3} + ....\infty1 + \dfrac{2}{3}.\dfrac{1}{2} + \dfrac{2.5}{3.6}(\dfrac{1}{2})^{2} + \dfrac{2.5.8}{3.6.9}(\dfrac{1}{2})^{3} + ....\infty

Find sum of 3+33+333+3333+3+33+333+3333+....... up to n terms.

Find the sum of the series (1^2+1)1!+(2^2+1)2!+(3^2+1)3!+......(n^2+1)n!(1^2+1)1!+(2^2+1)2!+(3^2+1)3!+......(n^2+1)n!

3+33+333+.....+3+33+333+.....+n terms.

The sum of the series \displaystyle \sum _{ k=1 }^{ 62 }{ \frac { 1 }{ \left( k+2 \right) \sqrt { k+1 } +\left( k+1 \right) \sqrt { k+2 } } } \displaystyle \sum _{ k=1 }^{ 62 }{ \frac { 1 }{ \left( k+2 \right) \sqrt { k+1 } +\left( k+1 \right) \sqrt { k+2 } } } .

How many three-digit numbers are there with no digit repeated?

Find the sum of the series 1.3^{2}+2.5^{2}+3.7^{2}+1.3^{2}+2.5^{2}+3.7^{2}+ to nn terms.

Solve:\frac{1}{{2.5}} + \frac{1}{{5.8}} + \frac{1}{{8.11}} + ...\frac{1}{{(3n - 1)(3n + 2)}}\frac{1}{{2.5}} + \frac{1}{{5.8}} + \frac{1}{{8.11}} + ...\frac{1}{{(3n - 1)(3n + 2)}}

Class 11 Commerce Maths Extra Questions

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Find the following sum.
$\sum\limits_{r = 1}^n {\left( {6{r^2} - 2r + 6} \right)}$

$1.2 + 2.3x + 3.4x^{2} + ....\infty (|x| < 1)$ .

Write the general term for the given series $\frac{1}{(n-1)},\frac{1}{3(n-3)},\frac{1}{5(n-5)},....$

Solve:

$\displaystyle\sum\limits_{r = 1}^n {\left( {6{r^2} - 2r + 6} \right)}$

The product of two consecutive positive numbers is $182$ . Find the numbers.

Evaluate $1+i^{2}+i^{4}+i^{6}+...+i^{2n}$ .

Sum the following series
$1 + \dfrac {3}{4} + \dfrac {7}{16} + \dfrac {15}{64} + \dfrac {31}{256} + ....$ to infinity.

Find value of the sum $\displaystyle \frac{1}{sin45^o sin 46^o} + \frac{1}{sin47^o sin 48^o} + \frac{1}{sin49^o sin 50^o} + ....... + \frac{1}{sin133^o sin 134^o}$

If $\displaystyle\frac{1}{1!10!}+\frac{1}{2!9!}+\frac{1}{3!10!}+...+\frac{1}{1!10!}=\frac{2}{k!}(2^{k-1}-1)$ then find the value of k.

If $\beta \, \neq \, 1$ be any nth root of unity then prove that $1 \, + \, 3\beta \, + \, 5\beta^2 \, + \, ..... \, + \, n \, terms \, = \, -\dfrac{2n}{1 \, - \, \beta}$

Let a sequence be defined by $a_1 = 1, a_2 = 1$ and, $a_n=a_{n-1} + a_{n - 2}$ for all $n > 2$ ,
find $\dfrac{a_{n + 1}}{a_n}$ for $n = 1, 2, 3, 4$ .

A sequence is defined by $a_n = n^3 - 6n^2 + 11n - 6$ . Show that the first three terms of the sequence are zero and all other terms are positive.

$\underset{n \rightarrow \infty}{lim} \dfrac{1^P + 2^P + 3^P + ....+ n^P}{n^{P+ 1}}$ equals-

Evaluate: $\sum_\limits{r=1}^n (3r-1)(r+1)$

Evaluate $\sum _{ k=1 }^{ 11 }{ \left( 2+{ 3 }^{ k } \right) }$

Solve :
$\dfrac {1}{1.2} + \dfrac {1}{2.3} + ...... + \dfrac {1}{n(n+1)} = ?$

Obtain $\sum\limits_{r = 2}^{10} {\left( {4{r^2} - 28r + 4a} \right)}$

Find the sum to $n$ terms of the series
$1.2^{2}+2.3^{2}+3.4^{2}+...$

Find the sum to n terms of the sequence, $8, 88, 888, 8888,....$ .

$\cfrac{1}{1.3.5}+\cfrac{1}{3.5.7}+....$ to $n$ terms.

Find the sum to $n$ terms of the series, whose $n$ th term is given by ${(2n-1)}^{2}$ .

Find the sum of series upto $n$ terms whose ${ n }^{ th }$ term is ${\left( {2n - 1} \right)^2}$

Find the sum of $1 + 5 + {5^2} + ....$ upto 8 terms.

Find the sum of the series $2 ^ { 2 } + 4 ^ { 2 } + 6 ^ { 2 } + \ldots + ( 2 n ) ^ { 2 }$

Show that $1^{2}+(1^{2}+2^{2})+(1^{2}+2^{2}+3^{2})+........$ upto $n$ terms $=\dfrac{n(n+1)^{2}(n+2)}{12}, \forall n \in N$

Find the sum of the series whose $n^{th}$ term is :
$2n^{3}+3n^{2}-1$

$\left( r+1 \right) ^{ 2 }+\left( r+2 \right) ^{ 2 }+....+{ n }^{ 2 }$

Find sum of $n$ terms of following series ?
$1 + 3 x + 5 x ^ { 2 } + 7 x ^ { 3 } + \dots$

For all real numbers $a, b$ and positive integer $n$ prove that:
$(a+b)^{n}=^{n}C_{0}a^{n}+^{n}C_{1}a^{n-1}b+^{n}C_{1}a^{n-2}b^{2}+..........+^{n}C_{n}b^{n}$ .

Prove that
$\dfrac {C_{1}}{C_{0}}+\dfrac {2C_{2}}{C_{1}}+\dfrac {3C_{3}}{C_{2}}+....+\dfrac {n.C_{n}}{C_{n-1}}=\dfrac {n(n+1)}{2}$

Write the first five terms of the sequence and obtain the corresponding series:
$a_{1}=a_{2}=2,a_{n}=a_{n-1}-1,n > 2$

Evaluate:
$\dfrac{1+\dfrac{1}{2!}+\dfrac{1}{3!}+.......}{1+\dfrac{1}{3!}+\dfrac{1}{5!}+.......}$ .

Find the coefficient of $x$ in the expansion of $\left( 1+\dfrac { x }{ 1! } +\dfrac { x^2 }{ 2! } +\dfrac { x^3 }{ 3! } +...+\dfrac { x^n }{ n! } \right)$

The sum, $\sum _{ n=1 }^{ 7 }{ \cfrac { n\left( n+1 \right) \left( 2n+1 \right) }{ 4 } }$ is equal to ________

Determine whether the following series is convergent or divergent
$\displaystyle \frac {1}{1^p}+\frac {1}{3^p}+\frac {1}{5^p}+\frac {1}{7^p}+.....$

Find the general term and the sum of $n$ terms of the series:
$2,5,12,31,86,......$

Determine whether the following series is convergent or divergent
$\displaystyle 1+ \frac {1}{2^2}+\frac {2^2}{3^3}+\frac {3^3}{4^4}+\frac {4^4}{5^5}....$

Determine whether the following series is convergent or divergent
$\displaystyle1 + \frac {2}{5}x+\frac {6}{9}x^2+\frac {14}{17}x^3+....+\frac {2^n-2}{2^n+1}x^{n-1}+....$

If $y=x+ x^2 + x^3 + ...\infty$ , where $|x| < 1$ , then prove that

$x= \left( \dfrac{y}{1+y} \right)$

Find the sum of the infinite series $1 + \dfrac{2}{3}.\dfrac{1}{2} + \dfrac{2.5}{3.6}(\dfrac{1}{2})^{2} + \dfrac{2.5.8}{3.6.9}(\dfrac{1}{2})^{3} + ....\infty$

Find sum of $3+33+333+3333+$ ....... up to n terms.

Find the sum of the series $(1^2+1)1!+(2^2+1)2!+(3^2+1)3!+......(n^2+1)n!$

$3+33+333+.....+$ n terms.

The sum of the series $\displaystyle \sum _{ k=1 }^{ 62 }{ \frac { 1 }{ \left( k+2 \right) \sqrt { k+1 } +\left( k+1 \right) \sqrt { k+2 } } }$ .

Find the sum of the series $1.3^{2}+2.5^{2}+3.7^{2}+$ to $n$ terms.

Solve:
$\frac{1}{{2.5}} + \frac{1}{{5.8}} + \frac{1}{{8.11}} + ...\frac{1}{{(3n - 1)(3n + 2)}}$