Question

The value of summation of
(n+1)^2/7^n is
n starts from 0 to infinity...

Answer 1

My ans is 343/180 ...bt there is no option as such

Answer 2

I am not in high school.

Answer 3

Is this your sum?

\[\sum_{n=0}^\infty \frac{(n+1)^2}{7^n}\]

Answer 4

Yep !

Answer 5

\[ \dfrac{1}{1-x}=\sum\limits_{n}x^n\]
\[ \dfrac{1}{(1-x)^2}=\sum\limits_{n}nx^{n-1}\]

\[ \dfrac{x}{(1-x)^2}=\sum\limits_{n}nx^{n}\]

differentiate again

Answer 6

Since your sum looks a lot like the geometric series -- something we know exactly how to evaluate -- it's a good idea to see if you can recreate your series from the geometric series.

So let's start here:

\[\frac{1}{1-x} = \sum_{n=0}^\infty x^n\]

Looks like we can plug in \(x = \tfrac{1}{7}\) and that is starting to look close, but we we want some \((n+1)\) factors, like what ganeshie is doing. :P

Answer 7

Well i did the same !!
@kainui there will be two agp's solving them i got this result

Answer 8

First is dividing by 7 and subtract it from S
Then again divide the term 6S/7 by 7 and subtract it from 6S/7
Finally we will have a gp of ratio 2/7

Answer 9

Well I guess I'll work it out and show my work so we can see what happened.

\[\frac{1}{1-x} = \sum_{n=0}^\infty x^n\]

Multiply both sides by x and then take the derivative:

\[ \frac{x}{1-x} = \sum_{n=0}^\infty x^{n+1}\]
\[\frac{d}{dx} \left( \frac{x}{1-x} \right) = \sum_{n=0}^\infty (n+1)x^{n}\]
\[ \frac{1}{(1-x)^2} = \sum_{n=0}^\infty (n+1)x^{n}\]
Repeat:

\[ \frac{d}{dx} \left( \frac{x}{(1-x)^2} \right) = \sum_{n=0}^\infty (n+1)^2x^{n}\]
\[ \frac{1+x}{(1-x)^3} = \sum_{n=0}^\infty (n+1)^2x^{n}\]
Now plug in \(x = \tfrac{1}{7}\) and hope it works out:

\[\frac{49}{27} = \sum_{n=0}^\infty (n+1)^27^{-n}\]

Answer 10

You want to work it w/o derivatives is it ?

Answer 11

Can that be in a new question?

Answer 12

Won't we be able to do if we try to solve the two agp's

Answer 13

I dunno, we can probably solve this some other way as well this just happened to be the way I saw to solve it first, I don't know what agp stands for sorry :X

Answer 14

Can I have some medals? :3

Answer 15

Arithmetic geometric progression

Answer 16

\[\begin{align}S &= \sum\limits_{n} \dfrac{n^2}{7^n} \\~\\\dfrac{1}{7}S &=\sum\limits_{n} \dfrac{n^2}{7^{n+1}}\\~\\ \dfrac{6}{7}S &=\sum\limits_{n}\dfrac{2n+1}{7^n} \end{align}\]

Answer 17

Correct again we will divide this by 7 and subtract it from 6s/7

Answer 18

Okk..i got the ans. .!!!

Answer 19

Sorry calculation mistake was being done by me...sorry guys....