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If S1, S2, and S are the sums of n terms, 2n terms and to infinity of a G.P. Then, find the value of S1(S1-S).

PS:Nothing is given about the common ratio.

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Of course no,this again comes from my test paper without any kind of explanation,except the the answer. – Quixotic Sep 28 '10 at 5:43
@Deb: You should state the source of the problem in the post. People are resistive to homework-like questions as one is supposed to do their own homework. – kennytm Sep 28 '10 at 6:33
@Debanjan: I mean you can just copy your first comment into the post in your next question (if any). – kennytm Sep 28 '10 at 8:24
@Debanjan: I suppose you have been asked if it is homework for earlier questions (hence your usage of word 'again'). Why don't you just mention that that is the case (from a test) and avoid getting questions like these ('is it homework')? In any case, why don't you also show some working? Test questions are like homework, in a way. – Aryabhata Sep 28 '10 at 16:24
@ Moron: The wike defination( and to my understanding homework is something which is to be assigned by my teacher and in case you didn't manage to do it he/she is there to help me doing it, where as I don't think test questions are since in some cases questions are not well defined and there is error in the solutions thanks to problem-setters.Besides you need a fast/tricky approach to get things done during the test. – Quixotic Sep 29 '10 at 4:58
up vote 4 down vote accepted

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.

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That's what I have done but let me ask you why are you assuming that r is less than 1 ? This satisfies the relation of-course.But in real time I can only spare a mint or so in this problem,so I guess the problem is not well defined ?! – Quixotic Sep 29 '10 at 4:52
I used formula (1) to evaluate the limit of $S_n$ as $n$ tends to $\infty$. This limits exists if and only if $|r|\lt 1$. – Américo Tavares Sep 29 '10 at 7:53

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.

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