Question

Find the sum of $n$ terms of the series
$(a+b)+(a^2+ab+b^2)+(a^3+a^{2}b+a{b}^2+b^3)+.....$ where $a\ne 1,b\ne 1anda\ne b$

• A. $\frac{1}{(a-b)}\left\{\frac{a^2(1-a^n)}{1-a}\right\}-\frac{1}{(a-b)}\left\{\frac{b^2(1-b^n)}{1-b}\right\}$
• B. $\frac{1}{(a-b)}\left\{\frac{a^2\left(a^{n-1}\right)}{a-1}\right\}-\frac{1}{(a-b)}\left\{\frac{b^2(b^n-1)}{b-1}\right\}$`
• C. $\frac{1}{(a-1)}\left\{\frac{a^2(1-a^n)}{1-b}\right\}+\frac{1}{(a-1)}\left\{\frac{b^2(1-b^n)}{1-b}\right\}$
• D. none of these

Answer 1

The correct option is A $\frac{1}{(a-b)}\left\{\frac{a^2(1-a^n)}{1-a}\right\}-\frac{1}{(a-b)}\left\{\frac{b^2(1-b^n)}{1-b}\right\}$

$S_n=(a+b)+(a^2+ab+b^2)+(a^3+a^{2}b+a{b}^2+b^3)+....$
$\Rightarrow (a-b)S_n=(a^2-b^2)+(a^3-b^3)+(a^4-b^4)+...$
$\Rightarrow (a-b)S_n=(a^2+a^3+a^4+...)-(b^2+b^3+b^4+...)$
$\Rightarrow (a-b)S_n=\frac{a^2(1-a^n)}{1-a}-\frac{b^2(1-b^n)}{1-b}$
$\therefore S_n=\frac{a^2(1-a^n)}{(1-a)(a-b)}-\frac{b^2(1-b^n)}{(1-b)(a-b)}$