Question

Can someone please help? I don't know what to do! Will fan and medal!

Answer 1

Let |r| < 1,
$$S = \sum_{k=0}^{\infty} r^k,$$
and
$$T = \sum_{k=0}^{\infty} k r^k.$$
The approach is to write T as a geometric series in terms of S and r.

Give a closed form expression for T in terms of r.

Answer 2

nice problem

Answer 3

What I got is,
T=(S-1)/(1-r)

(I won't show the work now, not to violate the policy)

Answer 4

that isn't correct :(

I think it needs to just be in terms of r.

Answer 5

it says "in terms of r and S"

Answer 6

But, S is just a geometric series, so that should be easy enough.

\(\color{#000000 }{ \displaystyle S=\sum_{k=0}^\infty r^k=\frac{1}{1-r} }\)

So,

\(\color{#000000}{{\rm T}=\dfrac{{\rm S}-1}{1-r}=\dfrac{\dfrac{1}{1-r}-\dfrac{1-r}{1-r}}{1-r}= \dfrac{\dfrac{r}{1-r}}{1-r}=?}\)

Answer 7

r/(r^2-2r+1) That is correct! Thanks!

Answer 8

Well, the point is rather, how to get to (S-1)/(1-r)

Answer 9

\(\color{#000000 }{ \displaystyle {\rm T}- {\rm S}=\sum_{k~=0}^{\infty}kr^k-\sum_{k=0}^{\infty}r^k }\)

\(\color{#000000 }{ \displaystyle {\rm T}- {\rm S}=\sum_{k~=0}^{\infty}(kr^k-r^k) }\)

\(\color{#000000 }{ \displaystyle {\rm T}- {\rm S}=\sum_{k~=0}^{\infty}r^k(k-1) }\)

\(\color{#000000 }{ \displaystyle {\rm T}- {\rm S}=r\sum_{k~=0}^{\infty}r^{k-1}(k-1) }\)

\(\color{#000000 }{ \displaystyle {\rm T}- {\rm S}=r\sum_{k~=-1}^{\infty}kr^{k} }\)

\(\color{#000000 }{ \displaystyle {\rm T}- {\rm S}=r\left(k_{(-1)}+\sum_{k~=0}^{\infty}kr^{k}\right) }\)

\(\color{#000000 }{ \displaystyle {\rm T}- {\rm S}=r\left(\frac{-1}{r}+\sum_{k~=0}^{\infty}kr^{k}\right) }\)

\(\color{#000000 }{ \displaystyle {\rm T}- {\rm S}={\rm T}r-1 }\)

\(\color{#000000 }{ \displaystyle {\rm T}-{\rm T}r= {\rm S}-1 }\)

\(\color{#000000 }{ \displaystyle {\rm T}(1-r)= {\rm S}-1 }\)

\(\color{#000000 }{ \displaystyle {\rm T}= ({\rm S}-1)/(1-r) }\)

this is how I got there