Question

Find the sum of the following series.
$(a^2-b^2).(a-b),\left(\frac{a-b}{a+b}\right),.............$ to $n$ terms.

Answer 1

Given series is,

$(a^2-b^2),(a-b),\left(\frac{a-b}{a+b}\right),...........$

Here, First term $=(a^2-b^2)$ and $r=\frac{1}{a+b}$

$\therefore S_n=\frac{a(1-r^n)}{1-r}$

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

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

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