Question

find the sum of the series

Answer 1

\[\sum_{n=2}^{\infty}n(n-1)x^n\]

Answer 2

i know what the answer is, i just dont know how they got it

Answer 3

the answer is \[\frac{ 2x^2 }{ (1-x)^3 }\]

Answer 4

i know we are differentiating...

Answer 5

yes, i know that much

Answer 6

and the sum before this one was \[\frac{ x }{ (1-x)^2 }\]

Answer 7

is the deriv, of x x^2? that doesnt seem right

Answer 8

is is that i need to use the quotient rule when i take the derivative?

Answer 9

Ok so notice that \[\frac{d}{dx}x^n=nx^{n-1}\]\[\frac{d}{dx}nx^n=n(n-1)x^{n-2}\]


\[\frac{d}{dx}\frac{1}{1-x}=\frac{1}{(1-x)^2}\]\[\frac{d}{dx}\frac{1}{(1-x)^2}=\frac{2}{(1-x)^3}\]

Then you just need a bit to manipulate the sum a little.

Answer 10

ok, yes. that is what i got too... where is the x^2 coming in at?

Answer 11

i kept getting \[\frac{ 2x }{ (1-x)^3 }\]

Answer 12

they say it is 2x^2

Answer 13

\[\sum_{n=2}^{\infty}n(n-1)x^{n-2}=\frac{2}{(1-x)^3}\\
=\sum_{n=2}^{\infty}n(n-1)x^{n}x^{-2}=\sum_{n=2}^{\infty}n(n-1)x^{n}\cdot \frac{1}{x^2}\]

Answer 14

you can multiply both sides by \(x^2\) to bring it to the right side.

Answer 15

oohhhhhh

Answer 16

ok! i see now!

Answer 17

thank you

Answer 18

:) your welcome