Geometric progression summation problem

Question

$n$ number of terms of a series are in geometric progression and their common ratio is $r$. Summation of the first $m$ terms gives $S_m$. Then prove that summation of products of two consecutive numbers in that series is $$\frac{r}{r+1} S_m S_{m-1} $$ when $n=2m$?

Answer 1

Let $(cr^n)_{n\geq 0}$ be a non-constant geometric progression then $$S_m=\sum_{n=0}^{m-1}cr^n=c\frac{1-r^{m}}{1-r}.$$ Now note that the "summation of products of two consecutive numbers in that series" is given by $$\sum_{n=0}^{m-1}(cr^n)(cr^{n+1})=c^2r\sum_{n=0}^{m-1}(r^2)^{n}=c^2r\frac{1-r^{2m}}{1-r^2}=\frac{r}{r+1}S_{1}S_{2m}.$$ Something is wrong in your statement.

Answer 2

The original geometric progression for $n=2m$ is: $$a,ar,ar^2,...,ar^{2m-2},ar^{2m-1}.$$ The sum of $m$ and $m-1$ terms: $$S_m=\frac{a(r^m-1)}{r-1}; S_{m-1}=\frac{a(r^{m-1}-1)}{r-1}.$$

The geometric progression formed by products of consecutive terms (overall $m$ terms) is: $$a^2r,a^2r^3,...,a^2r^{4m-3}.$$ The sum of $m$ terms: $$S_m^*=\frac{a^2r((r^2)^m-1)}{r^2-1}=\frac{r}{r+1}\cdot \frac{a(r^m-1)}{r-1}\cdot a(r^m+1)=\frac{r}{r+1}\cdot S_m\cdot a(r^m+1).$$