Question

If $S_1,S_2,S_3$ be respectively the sums of $n,2n,3n$ terms of a $G.P.$, then prove that
If the series $a_1,a_2,a_3,.a_n$ be a $G.P.$ of common ratio $r$, then prove that
$\frac{1}{{a_1}^m+{a_2}^m}+\frac{1}{{a_2}^m+{a_3}^m}+...+\frac{1}{a_{n-1}^m+a_{n-1}^m}=\frac{1-r^{m(1-n)}}{a_1^m(r^m-r^{-m})}$

Answer 1

$a_2=a_{1}r,a_3=a_{2}r,..,a_n=a_{n-1}r$
$S=\frac{1}{1+r^m}[\frac{1}{a_1^m}+\frac{1}{a_2^m}+.....\frac{1}{a_{n-1}^m}]=\frac{1}{1+r^m}.S_1$
Now
$S_1=[\frac{1}{a_1^m}+\frac{1}{{(a_1.r)}^m}+\frac{1}{{(a_1.r^2)}^m}+.....\frac{1}{{(a_1.r^{n-2})}^m}]$
$=\frac{1}{a_1^m}[\text{Sum of G.P}.of(n-1)\text{terms whose first term is 1 and common ration is}\frac{1}{r^m}]$
$\therefore S_1=\frac{1}{a_1^m}\left[\frac{1.[1-{\left(\frac{1}{r^m}\right)}^{n-1}]}{1-\frac{1}{r^m}}\right]=\frac{1}{a_1^m}\left[\frac{1-r^{-m(n-1)}}{1-r^{-m}}\right]$
$\therefore S=\frac{1}{1+r^m}.S_1=\frac{1}{a_1^m}\left[\frac{1-r^{m(1-n)}}{(1+r^m)(1-r^{-m})}\right]$