Imagine that the radius of convergence of $f(x)=\sum a_n x^n$ is infinity. We know that $ \int_{0}^4 f(x) dx =1$, find $\sum \frac{a_n n^2 4^n}{n+1}$. Imagine that the radius of convergence of $f(x)=\sum a_n x^n$ is infinity. We know that $ \int_{0}^4 f(x) dx =1$, find $\sum \frac{a_n n^2 4^n}{n+1}$.
I did some calculating and found out that $\frac{\int f(x)dx}{x} + f'(x) =  \sum \frac{a_n n^2 4^n}{n+1}$ First I got an integral of $f$, then a derivative, multiply by $x$,derivative, multiply by $x$.
But I don't know how to get the final result, and or if what I've done up to now is correct. Any ideas?
 A: Suppose $f(x) = a_0$. That is, $\forall n > 0, a_n = 0$. $\int_0^4f(x)\,dx = 1$ tells us that $a_0 = \frac 14$. And we easily get $$\sum_{n=0}^\infty \frac {n^2a_n4^n}{n+1} = 0$$
Now suppose $f(x) = a_1x$. Then $\int_0^4f(x)\,dx = 1$ tells us $a_1 = \frac 18$, and we get
$$\sum_{n=0}^\infty \frac {n^2a_n4^n}{n+1} = \frac 14$$
So there just isn't enough information given about $f$ for us to calculate a fixed value for the summation. The best we can do is to give an expression depending on $f$. I suggest doing the differentiations first.
$$xf'(x) = \sum_{n=0}^\infty na_nx^n\\x\frac{d}{dx}(xf'(x)) = \sum_{n=0}^\infty n^2a_nx^n\\xf'(x) + x^2f''(x) = \sum_{n=0}^\infty n^2a_nx^n$$
Now do the integration, with a division by $4$ to get the right exponent:
$$\frac14\int_0^4xf'(x) + x^2f''(x)\,dx = \frac 14\sum_{n=0}^\infty n^2a_n\int_0^4 x^n\,dx = \sum_{n=0}^\infty\frac{n^2a_n4^n}{n+1}$$
Integration by parts on the left yields $$\sum_{n=0}^\infty\frac{n^2a_n4^n}{n+1} = 4f'(4) -f(4) + \frac 14$$
