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A geometric series has common ratio $r$ where $|r|<1$. The sum of the first n terms is $S_n$ and the sum to infinity is $S$. Express $r$ in terms of $S_n$, $S$ and $n$, and prove that the sum of the first $2n$ terms is $\dfrac{[Sn(2S-Sn)]}{S}$.

I tried to use the formula $S_n=\dfrac{a(r^n -1)}{(r-1)}$ and $S_\infty=\frac{a}{(1-r)}$. But how to substitute the $a$ and $r$ so that the answer will be no $a$ and $r$?

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    No trace of your work on the subject... – 2017-01-15
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    Hint: eliminate $a$ between the formulas for $S_n$ and $S$ then you get $r$ in terms of $S_n,S,n\,$. – 2017-01-15
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    Use all the formula you know and show your work. – 2017-01-15
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    Did you want to write $Sn(2S-Sn)$? Or is it supposed to be $S_n(2S-S_n)$? – 2017-01-15

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The sum of a geometric series with first term $a$ and common ratio $r$ upto $n$ terms is given by $$S_n=\frac{a(1-r^{n})}{1-r}$$ The sum to infinity, $S$, is given by $$S=\lim_{n\to\infty}S_n=\frac{a}{1-r}$$ So, $$S_n=\frac{a}{1-r}\times(1-r^n)=S(1-r^n)$$ $$r^n=1-\frac{S_n}{S}$$ $$r=\left(1-\frac{S_n}{S}\right)^{\frac1n}$$

Also, $$S_{2n}=\frac{a(1-r^{2n})}{1-r}=S(1-r^{2n})$$ So, $$\frac{S_n(2S-S_n)}{S}=\frac{S(1-r^n)(2S-S(1-r^n))}{S}=S(1-r^n)(1+r^n)=S(1-r^{2n})=S_{2n}$$