From the question we have
![{\displaystyle \sum _{n=0}^{\infty }nx^{n+1}=x^{2}\sum _{n=0}^{\infty }nx^{n-1}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/24fbe7aaf56e2f55d69f4ecb92f945ff548217d8)
Therefore, it is enough to find an explicit expression of
.
For simplicity we denote
.
Observe that
. Using this,
can be written as
![{\displaystyle f(x)=\sum _{n=0}^{\infty }nx^{n-1}=\sum _{n=0}^{\infty }{\frac {d}{dx}}x^{n}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/cd0cea6535734ff72706924d8947208f798484e6)
Since the interval of convergence of the power series is
we can reverse the order of summation and derivative on this interval to get
![{\displaystyle f(x)={\frac {d}{dx}}\left(\sum _{n=0}^{\infty }x^{n}\right).}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/e5acfe47e30b67709cc2fab026d2371e286ff015)
By the hint (which can be easily obtained from the explicit expression of a geometric series), this implies that
![{\displaystyle f(x)={\frac {d}{dx}}\left({\frac {1}{1-x}}\right).}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/b731bcd082ed13d5de54a7e5b0ac07392d5ec87d)
Therefore, computing the derivative based on the chain rule and power rule, the explicit formula for
is given by
![{\displaystyle f(x)={\frac {d}{dx}}\left({\frac {1}{1-x}}\right)={\frac {1}{(1-x)^{2}}}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/c1356e744d85a697a1b0e59efcf09f4b608c5969)
Combining with the first equation, we get
Answer: