Question

How to show that `S_oo- S_n= (-1/3)^n` ,given, `S_oo=9/4` and `S_n=9/4(1-(-1/3)^n)`?

Answer 1

I think there's a `9/4` factor missing

Explanation:

Since you're given both `S_\infty` and `S_n`, in order to compute `S_\infty - S_n` you can simply subtract the two expressions:

`color(red)(S_\infty)-color(blue)(S_n) = color(red)(9/4) - color(blue)(9/4(1-(-1/3)^n)`

We can expand `S_n` as follows:

`9/4(1-(-1/3)^n) = 9/4 - 9/4\times(-1/3)^n`

So, we have

`S_\infty-S_n`
`= 9/4-(9/4 - 9/4\times(-1/3)^n)`
`= cancel(9/4)-cancel(9/4) + 9/4\times(-1/3)^n`

So,

`S_\infty-S_n =9/4\times(-1/3)^n`