Geometric Progression If  $S_1$, $S_2$ and $S$ are the sums of $n$ terms, $2n$ terms and to infinity of a G.P. Then, find the value of $S_1(S_1-S)$.
PS: Nothing is given about the common ratio.
 A: I change your notation from S1, S2 and S to $S_{n},S_{2n}$ and $S$.
The sum of $n$ terms of a geometric progression of ratio $r$
$u_{1},u_{2},\ldots ,u_{n}$
is given by
$S_{n}=u_{1}\times \dfrac{1-r^{n}}{1-r}\qquad (1)$.
Therefore the sum of $2n$ terms of the same progression is
$S_{2n}=u_{1}\times \dfrac{1-r^{2n}}{1-r}\qquad (2)$.
Assuming that the sum $S$ exists, it is given by 
$S=\lim S_{n}=u_{1}\times \dfrac{1}{1-r}\qquad (3)$.
Since the "answer is S(S1-S2)", we have to prove this identity 
$S_{n}(S_{n}-S)=S(S_{n}-S_{2n})\qquad (4).$
Plugging $(1)$, $(2)$ and $(3)$ into $(4)$ we have to prove the following equivalent algebraic identity:
$u_{1}\times \dfrac{1-r^{n}}{1-r}\left( u_{1}\times \dfrac{1-r^{n}}{1-r}%
-u_{1}\times \dfrac{1}{1-r}\right) $
$=u_{1}\times \dfrac{1}{1-r}\left( u_{1}\times \dfrac{1-r^{n}}{1-r}-u_{1}\times \dfrac{1-r^{2n}}{1-r}\right)
\qquad (5)$,
which, after simplifying $u_1$ and the denominator $1-r$, becomes: 
$\dfrac{1-r^{n}}{1}\left( \dfrac{1-r^{n}}{1}-\dfrac{1}{1}\right) =\left( \dfrac{%
1-r^{n}}{1}-\dfrac{1-r^{2n}}{1}\right) \qquad (6)$.
This is equivalent to
$\left( 1-r^{n}\right) \left( -r^{n}\right) =-r^{n}+r^{2n}\iff 0=0\qquad (7)$.
Given that $(7)$ is true, $(5)$ and $(4)$ are also true.
A: HINT $\quad\:$  In $\rm\  \ (1-X)\ (1-(1-X))\ =\ 1-X^2-(1-X)\ \ \ $ put $\rm\ \ \ X = x^n\ $
then multiply both sides by $\rm\  1/(1-x)^2\ =\ S/(1-x)\:.\ \ $  More generally one has
$\rm\ \ (1-x^a)\:(1-x^b)\ =\ (1-x^a) + (1-x^b) - (1-x^{a+b})$
$\rm\quad\quad\quad\ \Rightarrow\quad\quad S_a\ S_b\ =\ S\ (S_a + S_b - S_{a+b})\:,\quad S_n = \displaystyle\frac{1-x^n}{1-x},\quad S = S_\infty = \frac{1}{1-x}$
This generalizes to arbitrary products $\rm\: S_{a}\: S_b\: S_c\cdots S_k\:$ using the Inclusion–exclusion principle.
