Question

If $S_n$ denotes the sum of n terms of G.P. whose common ratio is r, then prove that sum of the products taken two and two together of this series is $\frac{r}{r+1}{S}_n.S_{n-1}$

Answer 1

$s_n=\frac{a(1-r{)}^n}{1-r},S_{n-1}=\frac{a(1-r^{n-1}}{1-r}$
Now we know that
$(x+y+z+.....{)}^2=\sum x^2+2\sum xy$
$\therefore 2\sum xy=(\sum x{)}^2-\sum x^2$
$\sum x^2=a^2+a^2{r}^2+a^2{r}^4+...nterms$
$=\frac{a^2(1-r^n{)}^2}{1-r^2}$
$2\sum xy=\frac{a^2(1-r^n{)}^2}{(1-r{)}^2}-\frac{a^2(1-r^{2n})}{1-r^2}$ from (1)
$\frac{a^2(1-r^n)}{(1-r)}[\frac{1-r^n}{1-r}-\frac{1+r^n}{1+r}]$
$=\frac{a^2(1-r^n)}{1-r}.\frac{2(r-r^n)}{(1-r)(1+r)}$
$=\frac{2r}{1+r}.\frac{a(1-r^n)}{1-r}.\frac{a(1-r^{n-1})}{1-r}$
$=\frac{2r}{1+r}{s}_n.S_{n-1}$
$\therefore sumxy=\frac{r}{1+r}{S}_n.S_{n-1}$