Arithmetico-geometric Series

A series is said to be an arithmetico geometric series if its each term is formed by multiplying the corresponding terms of an A.P and a G.P.

E.g. $1+2\mathrm{x}+3{\mathrm{x}}^2+4{\mathrm{x}}^3+$.

Here $1,2,3,4\dots$. are in A.P and $1,x,x^1,x^3\dots \dots$ are in G.P

Sum to $n$ terms

Let $S_n=a+(a+d)r+(a+2d)r^2+\dots \dots .(a+(n-1)d)r^{n-1}\dots \dots \dots (1)$

Multiply by $\mathrm{r}$ on both the sides

$\begin{array}{rl}& r{S}_n=ar+(a+d)r^2+\dots \dots ..(a+(n-1)d)r^n\dots \dots .(2)\\ & (1)-(2)\Rightarrow S_n(1-r)=a+(dr+d{r}^2+\dots ..d{r}^{n-1})-(a+(n-1)d)r^n\\ & (n-1)\text{terms}\end{array}$

$\begin{array}{rl}& a+dr\frac{(1-r^{n-1})}{1-r}-(a+(n-1)d)r^n\\ & S_n==\frac{a}{1-r}+\frac{dr(1-r^{n-1})}{(1-r{)}^2}-\frac{(a+(n-1)d)r^n}{1-r}\end{array}$

Sum to infinity

If $|\mathrm{r}|<1$ and $\mathrm{n}\to \mathrm{\infty }$, then $\underset{\mathrm{n}\to \mathrm{\infty }}{lim}{\mathrm{r}}^{\mathrm{n}}=0$

$\therefore \mathrm{S}=\frac{\mathrm{a}}{1-\mathrm{r}}+\frac{\mathrm{dr}}{(1-\mathrm{r}{)}^2}$

Note: If we take the first term of a G.P to be b, then

${\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{ab}}{1-\mathrm{r}}+\frac{\mathrm{dbr}(1-{\mathrm{r}}^{\mathrm{n}-1})}{(1-\mathrm{r}{)}^2}-\frac{(\mathrm{a}+(\mathrm{n}-1)\mathrm{d}){\mathrm{br}}^{\mathrm{n}}}{1-\mathrm{r}}$

If $|\mathrm{r}|<1$, then sum to infinity, $S=\frac{\mathrm{ab}}{1-\mathrm{r}}+\frac{\mathrm{dbr}}{(1-\mathrm{r}{)}^2}$

Use of Natural numbers

$1$ Let ${\mathrm{S}}_{\mathrm{r}}=1^{\mathrm{r}}+2^{\mathrm{r}}+3^{\mathrm{r}}+$ $\dots \dots \dots .+{\mathrm{n}}^{\mathrm{r}}$, then

(i) ${\mathrm{S}}_1=1+2+3+\dots \dots \dots \dots \dots +\mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}$

(ii) ${\mathrm{S}}_2=1^2+2^2+3^3+\dots \dots \dots \dots \dots \dots +\dots \dots +{\mathrm{n}}^2=\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}$

(iii) ${\mathrm{S}}_3=1^3+2^3+3^3+\dots \dots \dots \dots \dots \dots \dots +{\mathrm{n}}^3=\frac{{\mathrm{n}}^2(\mathrm{n}+1{)}^2}{4}={\mathrm{S}}_1{}^2$

(iv) ${\mathrm{S}}_4=1^4+2^4+3^4+\dots \dots \dots \dots \dots \dots +{\mathrm{n}}^4=\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)(3{\mathrm{n}}^2+3\mathrm{n}-1)}{30}=\frac{{\mathrm{S}}_2}{5}(6{\mathrm{S}}_1-1)$

(iv) ${\mathrm{S}}_5=1^5+2^5+3^5+\dots \dots \dots \dots \dots \dots +{\mathrm{n}}^5=\frac{{\mathrm{n}}^2(\mathrm{n}+1{)}^2(2{\mathrm{n}}^2+2\mathrm{n}-1)}{12}=\frac{1}{3}{\mathrm{S}}_1{}^2(4{\mathrm{S}}_1-1)$

$21+3+5+$ ……….to $\mathrm{n}$ terms $={\mathrm{n}}^2$

$31^2+3^2+5^2+$……….to $\mathrm{n}$ terms $=\frac{\mathrm{n}(4{\mathrm{n}}^2-1)}{3}$

$41^3+3^3+5^3+$………..to $\mathrm{n}$ terms $={\mathrm{n}}^2(2{\mathrm{n}}^2-1)$

$51-1+1-$ ………..to $\mathrm{n}$ terms $=\frac{1-(-1{)}^{\mathrm{n}}}{2}$

$61-2+3-$ ………..to $\mathrm{n}$ terms $=\frac{1-(-1{)}^{\mathrm{n}}(2\mathrm{n}+1)}{4}$

$71^2-2^2+3^2-$ ………..to $\mathrm{n}$ terms $=\frac{(-1{)}^{\mathrm{n}-1}\mathrm{n}(\mathrm{n}+1)}{2}=(-1{)}^{\mathrm{n}-1}{\mathrm{S}}_1$

$81^3-2^3+3^3-$ ………..to $n$ terms $=\frac{(-1{)}^{n-1}(4n^3+6n^2-1)-1}{8}$

Application

If ${\mathrm{n}}^{\text{th }}$ term of a sequence is given by ${\mathrm{T}}_{\mathrm{n}}={\mathrm{an}}^3+{\mathrm{bn}}^2+\mathrm{cn}+\mathrm{d}$, where $\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\in \mathrm{R}$, then

$\begin{array}{rl}& {\mathrm{S}}_{\mathrm{n}}=\sum {\mathrm{T}}_{\mathrm{n}}={\mathrm{T}}_1+{\mathrm{T}}_2+\dots \dots \dots \dots \dots +{\mathrm{T}}_{\mathrm{n}}\\ & =\mathrm{a}\sum {\mathrm{n}}^3+\mathrm{b}\sum {\mathrm{n}}^2+\mathrm{c}\sum \mathrm{n}+\mathrm{d}\sum 1\end{array}$

Note: If ${\mathrm{T}}_{\mathrm{n}}$ is expressible as product of $\mathrm{m}$ consecutive numbers beginning with $\mathrm{n}$, i.e. $T_n=n(n+1)(n+2)\dots (n+m-1)$ then

${\mathrm{S}}_{\mathrm{n}}=\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)\dots (\mathrm{n}+\mathrm{m}-1)\left(\frac{\mathrm{n}+\mathrm{m}}{\mathrm{m}+1}\right)$

Eg. If ${\mathrm{T}}_{\mathrm{n}}=\mathrm{n}$, then ${\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}$ $($ Here $\mathrm{m}=1)$

E.g. $T_n=n(n+1)$, then $S_n=\frac{n(n+1)(n+2)}{3}$ $($ Here $\mathrm{m}=2)$

Method of differences

If the differences of successive terms of a series are in A.P. or G.P., we can find $T_n$ as follows

(a) Denote ${\mathrm{n}}^{\text{th }}$ term and the sum up to $n$ terms by $T_n\mathrm{\&}{S}_n$ respectively

(b) Rewrite the given series with each term shifted by one place to the right

(c) Subtracting the above two forms of the series, find $T_n$.

(d) Apply $S_n=\sum T_n$.

Note: Instead of determining the ${\mathrm{n}}^{\text{th }}$ item of a series by the method of difference, we can use the following steps to obtain the same

(i) If the differences ${\mathrm{T}}_2-{\mathrm{T}}_1,{\mathrm{T}}_3-{\mathrm{T}}_2$ ${\mathrm{T}}_{\mathrm{n}}={\mathrm{an}}^2+\mathrm{bn}+\mathrm{c},\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{R}$

Determine $a,b,c$ by putting $n=1,2,3$ and equating them with the values of corresponding terms of the given series.

(ii) If the differences $T_2-T_1,T_3-T_2,\dots \dots \dots \dots$…tc are in G.P, with common ratio $\mathrm{r}$, then take $T_n=$ $a^{n-1}+bn+c,a,b,c\in R$

Determine $\mathrm{a},\mathrm{b},\mathrm{c}$ by putting $\mathrm{n}=1,2,3$ and equating them with the values of corresponding terms of the given series.

(iii) If the differences of the differences computed in step (i) are in A.P, then take $T_n=a^3+$ ${\mathrm{bn}}^2+\mathrm{cn}+\mathrm{d}$

Determine $a,b$, by putting $n=1,2,3,4$ and equating them with the values of corresponding terms of the given series.

(iv) If the differences of differences computed in step (i) are in G.P with common ratio $\mathrm{r}$, then take ${\mathrm{T}}_{\mathrm{n}}={\mathrm{ar}}^{\mathrm{n}-1}+{\mathrm{bn}}^2+\mathrm{cn}+\mathrm{d}$

Determine $a,b$, by putting $n=1,2,3,4$ and equating them with the values of corresponding terms of the given series.

Summation by " $\sum$ " (sigma) operator

i. $\sum _{\mathrm{r}=1}^{\mathrm{n}}{\mathrm{T}}_{\mathrm{r}}={\mathrm{T}}_1+{\mathrm{T}}_2+\dots \dots .+{\mathrm{T}}_{\mathrm{n}}$

ii. $\sum _{\mathrm{r}=1}^{\mathrm{n}}1=1+1+1+\dots \dots +\mathrm{n}$ times $=\mathrm{n}$

iii. $\sum _{\mathrm{r}=1}^{\mathrm{n}}\mathrm{k}{\mathrm{T}}_{\mathrm{r}}=\mathrm{k}\sum _{\mathrm{r}=1}^{\mathrm{n}}{\mathrm{T}}_{\mathrm{r}};\mathrm{k}$ is a constant

iv. $\sum _{r=1}^n(T_r\pm T_r^1)=\sum _{r=1}^n{T}_r\pm \sum _{r=1}^n{T}_r^1$

v. $\sum _{j=1}^n\sum _{i=1}^n{T}_i{T}_j=\left(\sum _{i=1}^n{T}_i\right)\left(\sum _{j=1}^n{T}_j\right)$

(Note that $\mathrm{i}\mathrm{\&}\mathrm{j}$ are independent here)

vi. Now consider $\sum _{0\le i<j\le n}{f}_n(\mathrm{i})\times \mathrm{f}(\mathrm{j})$ Here three types of terms occur, for which $\mathrm{i}<\mathrm{j},\mathrm{i}>\mathrm{j}$ and $i=j$. Also note that the sum of terms when $i<j$ equal to the sum of the terms when $i>j$ if $f(i)$ and $f(j)$ are symmetrical. In such case,

$\begin{array}{rl}& \sum _{i=0}^n\sum _{j=0}^{n}f(i)f(j)=\sum _{0\le j<i\le n}\sum _{i}f(i)f(j)+\sum _{0\le i<j\le n}\sum _{i=j}f(i)f(j)+\sum _i\sum _f(i)f(j)\\ & =2\sum _{0\le i<j\le n}\sum _{f}f(i)f(j)+\sum _{i=j}\sum f(i)f(j)\\ & \Rightarrow \sum _{0\le i<j\le n}f(i)f(j)=\frac{\sum _{i=0}^n\sum _{j=0}^{n}f(i)f(j)-\sum _i\sum _{=j}f(i)f(j)}{2}\end{array}$

When $f(i)$ and $f(j)$ are not symmetrical, the sum can be obtained by listing all the terms.

Example: 1 $\sum _{\mathrm{i}=1}^4\sum _{\mathrm{j}=1}^3\mathrm{ij}=\sum \mathrm{i}(1+2+3)=\sum (\mathrm{i}+2\mathrm{i}+3\mathrm{i})$

$(1+2+3)+(2+4+6)+(3+6+9)+(4+8+12)=6+12+18+24=60$

Also, $\sum _{\mathrm{i}=1}^4\sum _{\mathrm{j}=1}^3\mathrm{i}=\sum _{\mathrm{i}=1}^4\mathrm{i}\sum _{\mathrm{j}=1}^3\mathrm{j}=\frac{4\mathrm{x}5}{2}\mathrm{x}\frac{3\times 4}{2}=60$ (Since $\mathrm{i}\mathrm{\&}\mathrm{j}$ are independent)

Note that $2\sum _{\mathrm{i}<\mathrm{j}=1}^{\mathrm{n}}{\mathrm{a}}_{\mathrm{i}}{\mathrm{a}}_{\mathrm{j}}={({\mathrm{a}}_1+{\mathrm{a}}_2+\dots ..+{\mathrm{a}}_{\mathrm{n}})}^2-({\mathrm{a}}_1^2+{\mathrm{a}}_2^2+\dots ..+{\mathrm{a}}_{\mathrm{n}}^2)$

Example: 2 $\sum \sum _{0\le i\mathrm{j}j\mathrm{n}}1$

$\begin{array}{rl}& =\frac{\sum _{i=1}^n\sum _{j=1}^{n}1-\sum _{i=j}\sum 1}{2}\\ & =\frac{\left(\sum _{i=1}^{n}1\right)\left(\sum _{j=1}^{n}1\right)-\sum _{j=1}^{n}1}{2}\\ & =\frac{n\cdot n-n}{2}=\frac{n(n-1)}{2}={}^n{C}_2\end{array}$

Example: 3 Consider $\sum _{\mathrm{i}=1}^{\mathrm{n}}\sum _{\mathrm{j}=1}^{\mathrm{n}}\mathrm{ij}$

There are 3 types of terms in this summation,

i. Those terms when $\mathrm{i}<\mathrm{j}$ (upper triangle)

ii. Those terms when $\mathrm{i}>\mathrm{j}$ (lower triangle)

iii. Those terms when $\mathrm{i}=\mathrm{j}$ (diagonal) It is shown in the diagram

$\sum _{\mathrm{i}=1}^{\mathrm{n}}\sum _{\mathrm{j}=1}^{\mathrm{n}}\mathrm{ij}=$ sum of terms in upper triangle + sum of terms in lower triangle + sum of terms in diagonal.

$\begin{array}{rl}& \sum _{\mathrm{i}=1}^{\mathrm{n}}\sum _{\mathrm{j}=1}^{\mathrm{n}}\mathrm{ij}=2\sum _{0\le i\le j\le \mathrm{n}}\sum _{\mathrm{i}j}\mathrm{ij}+\sum _{\mathrm{i}=\mathrm{j}}\sum _{\mathrm{ij}}(\because \text{ sum of terms in upper }+\text{ lower tringles are same })\\ & \sum _{0\le i\le j\le n}\sum _{\mathrm{ij}}=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}\sum _{\mathrm{j}=1}^{\mathrm{n}}\mathrm{ij}-\sum _{\mathrm{i}=\mathrm{j}}\sum \mathrm{ij}}{2}=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}\sum _{\mathrm{j}=1}^{\mathrm{n}}\mathrm{i}-\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{i}}^2}{2}\\ & =\frac{\frac{\mathrm{n}(\mathrm{n}+1)}{2}\cdot \frac{\mathrm{n}(\mathrm{n}+1)}{2}-\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}}{2}\\ & =\frac{\mathrm{n}({\mathrm{n}}^2-1)(3\mathrm{n}+2)}{24}\end{array}$

Solved examples

1. Find sum of the series to $n$ terms

2. $3^{\frac{1}{4}}\times 9^{\frac{1}{8}}\times {27}^{\frac{1}{16}}\times \dots \dots .$. to $\mathrm{\infty }=$

(a). $3$

(b). $9$

(c). $\frac{1}{3}$

(d). none of these

3. $\sum _{\mathrm{m}=1}^{\mathrm{\infty }}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{m}}^2\mathrm{n}}{3^{\mathrm{m}}(\mathrm{n}\cdot 3^{\mathrm{m}}+\mathrm{m}\cdot 3^{\mathrm{n}})}=$

(a). $\frac{16}{9}$

(b). $\frac{9}{16}$

(c). $1$

(d). none of these

4. Sum to $n$ terms of the series $1^2-2^2+3^2-4^2+$ is

5. Find the sum of all possible products of first $n$ natural numbers taken two by two

Solution: $\sum _{1\le i\le j\le n}{\mathrm{x}}_{\mathrm{i}}{\mathrm{x}}_{\mathrm{j}}=\frac{1}{2}\{\sum _{\mathrm{i}=1}^{\mathrm{n}}\sum _{\mathrm{j}=1}^{\mathrm{n}}\mathrm{ij}-\sum _{\mathrm{i}=\mathrm{j}}\sum \mathrm{ij}\}$

$\begin{array}{rl}& =\frac{1}{2}\{{\left(\frac{n(n+1)}{2}\right)}^2-\frac{n(n+1)(2n+1)}{6}\}\\ & =\frac{1}{24}n(n+1)(n-1)(3n+2)\end{array}$

6. Find sum to $\mathrm{n}$ terms of $\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+\dots \dots \dots$

7. $\frac{3}{4}+\frac{5}{36}+\frac{7}{144}+\frac{9}{400}+$ …….to $\mathrm{\infty }$ is

(a). 2

(b). 1

(c). 3

(d). none of these

8. Find the $n^{\text{th }}$ term of the following series

i. $3+7+13+21+\dots \dots$

ii. $5+7+13+31+85+\dots \dots .$

$V_n$ method

1. If $S_n=\frac{1}{a_1{a}_2\dots \dots .a_r}+\frac{1}{a_2{a}_3\dots \dots .a_{r+1}}+\dots \dots ..+\frac{1}{a_n{a}_{n+1}{a}_n\dots \dots a_{n+r-1}}$

$T_n=\frac{1}{a_n{a}_{n+1}\dots \dots a_{n+r-1}},V_n=\frac{1}{a_{n+1}\dots \dots a_{n+r-1}}$ (leave 1st term in denominator of $T_n$ )

$V_n-V_{n-1}=\frac{1}{a_{n+1}{a}_{n+2}\dots \dots a_{n+r-1}}-\frac{1}{a_n{a}_{n+1}\dots \dots a_{n+r-2}}$

$=\frac{1}{a_n{a}_{n+1}\dots \dots a_{n+r-1}}(a_n-a_{n+r-1})$

${\mathrm{V}}_{\mathrm{n}}-{\mathrm{V}}_{\mathrm{n}-1}={\mathrm{T}}_{\mathrm{n}}(1-\mathrm{r})\mathrm{d}$

${\mathrm{T}}_{\mathrm{n}}=\frac{-1}{\mathrm{d}(\mathrm{r}-1)}({\mathrm{V}}_{\mathrm{n}}-{\mathrm{V}}_{\mathrm{n}-1})\Rightarrow {\mathrm{S}}_{\mathrm{n}}=\frac{-1}{\mathrm{d}(\mathrm{r}-1)}({\mathrm{V}}_{\mathrm{n}}-{\mathrm{V}}_0)$

E.g $\frac{1}{1.2}+\frac{1}{1.3}+\dots \dots \dots +\frac{1}{n(n+1)}=\frac{n}{n+1}$

$\begin{array}{rl}& \frac{1}{1.2.3}+\frac{1}{2.3.4}+\dots \dots \dots \dots \dots +\frac{1}{n(n+1)(n+2)}=\frac{1}{2}(\frac{1}{1.2}-\frac{1}{(n+1)(n+2)})\\ & \frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\dots \dots \dots \dots \dots +\frac{1}{n(n+1)(n+2)(n+3)}=\frac{1}{3}(\frac{1}{1.2.3}-\frac{1}{(n+1)(n+2)(n+3)})\end{array}$

2. If $S_n=a_1{a}_2\dots \dots a_r+a_2{a}_3\dots \dots a_{r+1}+\dots \dots +a_n{a}_{n+1}\dots \dots a_{n+r-1}$

$\begin{array}{rl}& T_n=a_n{a}_{n+1}\dots \dots .a_{n+r-1}{V}_n=a_n{a}_{n+1}\dots \dots .a_{n+r-1}{a}_{n-r}\\ & V_n-V_{n-1}=a_n{a}_{n+1}\dots .a_{n+r}-a_{n-1}{a}_n\dots \dots .a_{n+r-1}\\ & =a_n{a}_{n+1}\dots .a_{n+r-1}(a_{n+r}-a_{n-1})\\ & =T_n(r+1)d\\ & T_n=\frac{1}{d(r+1)}(V_n-V_{n-1})\Rightarrow S_n=\frac{1}{d(r+1)}(V_n-V_0)\\ & 1.2+2.3+\dots ..+n(n+1)=n(n+1)\frac{(n+2)}{3}\end{array}$

$1\cdot 2\cdot 3\cdot 4+2\cdot 3\cdot 4\cdot 5+\dots \dots \dots +n(n+1)(n+2)(n+3)=n(n+1)(n+2)(n+3)\frac{(n+4)}{5}$

Alternative method

1. $\mathrm{S}=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\dots \dots \dots \dots +\frac{1}{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}$

$\begin{array}{rl}& =\frac{1}{2}\{\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\dots \dots \dots \dots +\frac{(n+2)-n}{n(n+1)(n+2)}\}\\ & =\frac{1}{2}\{(\frac{1}{1.2}-\frac{1}{2.3})+(\frac{1}{2.3}-\frac{1}{3.4})+\dots \dots +(\frac{1}{\mathrm{n}(\mathrm{n}+1)}-\frac{1}{(\mathrm{n}+1)(\mathrm{n}+2)})\}\\ & =\frac{1}{2}\{\frac{1}{1.2}-\frac{1}{(\mathrm{n}+1)(\mathrm{n}+2)}\}\end{array}$

2. $1.2.3+2.3.4+\dots \dots +n(\mathrm{n}+1)(\mathrm{n}+2)=\frac{(\mathrm{n})(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)}{(3+1)}$

$=\frac{{\mathrm{t}}_{\mathrm{n}}(\mathrm{n}+3)}{3+1}$

i.e. Sum of the product of $m$ consecutive natural $S_n=\frac{T_n(n+m)}{m+1}$

Practice questions

1. If $R\subset (0,\pi )$ denote the set of values which satisfies the equation

$2^{1+|\cos x|+|{\cos }^{2}x|+|{\cos }^{3}x|+\dots \dots }=4$, then $R=$

(a). $\left\{\frac{-\pi }{3}\right\}$

(b). $\{\frac{\pi }{3},\frac{2\pi }{3}\}$

(c). $\{\frac{-\pi }{3},\frac{2\pi }{3}\}$

(d). $\{\frac{\pi }{3},\frac{-2\pi }{3}\}$

2. Find the value of the expression $\sum _{\mathrm{i}=1}^{\mathrm{n}}\sum _{\mathrm{j}=1}^{\mathrm{i}}\sum _{\mathrm{k}=1}^{\mathrm{j}}1$

3. If in a series ${\mathrm{t}}_{\mathrm{n}}=\frac{\mathrm{n}}{(\mathrm{n}+1)!}$, Then $\sum _{\mathrm{n}=1}^{20}{\mathrm{t}}_{\mathrm{n}}$ is equal to

(a). $\frac{20!-1}{20!}$

(b). $\frac{21!-1}{21!}$

(c). $\frac{1}{2(n-1)!}$

(d). none of these

4. The value of $(0.2{)}^{{\log }_{\sqrt{5}}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\dots \dots \mathrm{\infty })}$ is

(a). $1$

(b). $2$

(c). $\frac{1}{2}$

(d). $4$

5. $\underset{n\to \mathrm{\infty }}{lim}(1+3^{-1})(1+3^{-2})(1+3^{-4})(1+3^{-8})\dots \dots ..(1+3^{-2^n})$ is equal to

(a). $1$

(b). $\frac{1}{2}$

(c). $\frac{3}{2}$

(d). none of these

6. $\underset{\mathrm{n}\to \mathrm{\infty }}{lim}\frac{1.2.3+2.3.4+3.4.5+\dots \dots \dots ...\mathrm{upto}\mathrm{n}\text{ terms }}{\mathrm{n}(1.2+2.3+3.4+\dots \dots \dots \text{. }}$ is equal to $\mathrm{n}$ terms $)$

(a). $\frac{3}{4}$

(b). $\frac{1}{4}$

(c). $\frac{1}{2}$

(d). $\frac{5}{4}$

7. Sum to infinite terms of the series ${\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{2}{9}+{\tan }^{-1}\frac{1}{8}+{\tan }^{-1}\frac{2}{25}+{\tan }^{-1}\frac{1}{18}+\dots \dots$.

(a). ${\tan }^{-1}3$

(b). ${\cot }^{-1}\frac{1}{3}$

(c). ${\tan }^{-1}\frac{1}{3}$

(d). ${\cot }^{-1}3$

8. If $f(x)=a_0+a_{1}x+a_2{x}^2+\dots \dots +a_n{x}^n+\dots \dots$. and $\frac{f(x)}{1-x}=b_0+b_{1}x+b_2{x}^2+\dots \dots .+b_n{x}^n+\dots \dots$. If $a_0=1$ and ${\mathrm{b}}_1=3$ and ${\mathrm{b}}_{10}={\mathrm{k}}^{11}-1$, and ${\mathrm{a}}_0,{\mathrm{a}}_1,{\mathrm{a}}_2,\dots .$. are in G.P then $\mathrm{k}$ is