c2 Geometrics Series

My question is:

Starting from first principles, prove that the sum of the first n terms of a geometric progression whose first term is a and whose common ratio is r

(where

r\neq 1

) is

S_{n} = \frac{a(1-r^{n})}{1-r}

a)Show that

\frac{S_{3n}-S_{2n}}{S_{n}}= r^{2n}

b)Given that

r= \frac{1}{2},

find

\sum_{n=1}^{\infty } = \frac{S_{3n}-S_{2n}}{S_{n}}

My answer for part a;

\frac{S_{3n}-S_{2n}}{S_{n}}= r^{2n}

LHS=RHS

LHS= \frac{S_{3n}-S_{2n}}{S_{n}}=\frac{\frac{a(1-r^{3n})}{1-r} -\frac{a(1-r^{2n})}{1-r}}{\frac{a(1-r^{n})}{1-r}}

\frac{ar^{2n}-ar^{3n}}{1-r}\times \frac{1-r}{a(1-r^{n})}

\frac{ar^{2n}(1-r^{n})}{a(1-r^{n})} = r^{2n}\equiv RHS

Please help me with part b
Thank you

Reply 1

mSayed

that's what I can get, hope it will help :smile:

Reply 2

bigmansouf

Original post by mSayed

that's what I can get, hope it will help :smile:

Thank you very much for your help

I forgot that since

LHS \equiv RHS

I could replace

\frac{S_{3n}-S_{2n}}{S_{n}}

with

r^{2n}

thank you very much

unfortunately I cant rep you since I am not allow to rep you twice in 24 hours

Reply 3

mSayed

Original post by bigmansouf

Thank you very much for your help

I forgot that since

LHS \equiv RHS

I could replace

\frac{S_{3n}-S_{2n}}{S_{n}}

with

r^{2n}

thank you very much

unfortunately I cant rep you since I am not allow to rep you twice in 24 hours

It doesn't matter, the important thing is that you got it now. :wink:

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c2 Geometrics Series

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