My question looks like this:
f(x)= 5/ (1-5x)^2
Find
C0=
c1=
C2=
C3=
C4=
Any help is much appreciated.
If $\displaystyle - 1 < \xi < 1$ then $\displaystyle \frac{1}{1-\xi } = \sum_{n=0}^{\infty} \xi^n$.
Term-by-term differenciation implies,
$\displaystyle \frac{1}{(1-\xi)^2} = \sum_{n=0}^{\infty}(n+1)\xi ^n$.
Thus if $\displaystyle -1 < 5x < 1 \implies -\frac15 < x <\frac15$ then,
$\displaystyle \frac{1}{(1-5x)^2} = \sum_{n=0}^{\infty}5^n(n+1)x^n$
Thus,
$\displaystyle \frac{5}{(1-5x)^2} = \sum_{n=0}^{\infty} 5^{n+1} (n+1)x^n$.
Its Taylor series is $\displaystyle \sum_{n=0}^{\infty} \frac{f^{(n)}(0)x^n}{n!}$.
This means, $\displaystyle f^{(n)}(0) = (n+1)! 5^{n+1} $.