Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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
1  
@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
1  
@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
1  
@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
1  
@ Moron: The wike defination(en.wikipedia.org/wiki/Homework) 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

2 Answers 2

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.

share|improve this answer
    
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 $|r|\lt 1$. en.wikipedia.org/wiki/Geometric_series –  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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.