Question

Find the sum of n terms of the series $(a+b)+(a^2+2b)+(a^3+3b)+.....$

Answer 1

Let $S=(a+b)+(a^2+2b)+.....+(a^n+nb)$

$=(a+a^2+.....+a^n)+(b+2b+.....+nb)$

Let $S=S_1+S_2.....\left(1\right)$

whereas,

$S_1=(a+a^2+.....+a^n)$

$S_2=(b+2b+.....+nb)$

Clearly $S_1$ is a series in G.P. with first term '$a$' and common ratio also '$a$' while $S_2$ i a series in A.P. with first term '$b$' and common ratio also '$b$'.

Therefore,

Case I:- If $a<1$

$S_1=\frac{a(1-a^n)}{1-a}$

$S_2=\frac{n}{2}[2b+(n-1)b]$

Substituting these values in ${eq}^n\left(1\right)$, we get

$S=\frac{a(1-a^n)}{1-a}+\frac{n}{2}[2b+(n-1)b]$

Case II:- If $a>1$

$S_1=\frac{a(a^n-1)}{a-1}$

$S_2=\frac{n}{2}[2b+(n-1)b]$

Substituting these values in ${eq}^n\left(1\right)$, we get

$S=\frac{a(a^n-1)}{a-1}+\frac{n}{2}[2b+(n-1)b]$

Hence the sum of $n$ terms of the series $(a+b)+(a^2+2b)+.....+(a^n+nb)$ will be-

$s=\{\begin{array}{c}\frac{a(1-a^n)}{1-a}+\frac{n}{2}[2b+(n-1)b];\text{if}a<1\\ \frac{a(a^n-1)}{a-1}+\frac{n}{2}[2b+(n-1)b];\text{if}a>1\end{array}$