Compute $\sum_{k=1}^{n+3} {(k+1)^2}\binom {n}{k}x^{k+1}$

Question

Compute $\sum_{k=1}^{n+3} {(k+1)^2}\binom {n}{k}x^{k+1}$

Well, I believe I have somehow to get rid of $x$, maybe differentation. anyways, look at my stupidity: (Help is appreciated)

Answer 1

You could absorb one of the factors $(k+1)$ into $x$ by writing the expression as the derivative of some other expression with respect to $x,$ if only the power of $x$ were $k$ instead of $k+1$. So bring one factor $x$ to the left and rewrite your expression as

$$x\frac{d}{dx}\sum_{k=1}^n(k+1)\left({n\atop k}\right)x^{k+1}$$

and repeating the same trick

$$...=x\frac{d}{dx}x\frac{d}{dx}\sum_{k=1}^n\left({n\atop k}\right)x^{k+1}$$

now the binomial formula applies if we bring one more $x$ outside the sum

$$ ...=x\frac{d}{dx}x\frac{d}{dx}x\sum_{k=1}^n\left({n\atop k}\right)x^{k} =x\frac{d}{dx}x\frac{d}{dx}x((x+1)^n-1) $$

Answer 2

\begin{align} f(x)&=\sum_{k=1}^{n+3} {(k+1)^2}\binom {n}{k}x^{k+1}\\ &=\sum_{k=1}^{n} {(k+1)^2}\binom {n}{k}x^{k+1}\\ \frac{f(x)}x&=\sum_{k=1}^{n} {(k+1)^2}\binom {n}{k}x^{k}\\ \int \frac{f(x)}xdx&=\sum_{k=1}^{n} {(k+1)}\binom {n}{k}x^{k+1}\\ \frac1x\int \frac{f(x)}xdx&=\sum_{k=1}^{n} {(k+1)}\binom {n}{k}x^{k}\\ \int\left(\frac1x\int \frac{f(x)}xdx\right)dx&=\sum_{k=1}^{n} {}\binom {n}{k}x^{k+1}\\ &=x\left((1+x)^n-1\right)\\ \frac1x\int \frac{f(x)}xdx&=\left((1+x)^n-1\right)+nx(1+x)^{n-1}-1\\ \int \frac{f(x)}xdx&=x\left((1+x)^n-1\right)+nx^2(1+x)^{n-1}-x\\ \frac{f(x)}x&=x\left((1+x)^n-1\right)+nx^2(1+x)^{n-1}-x+2nx(1+x)^{n-1}+(n-1)nx^2(1+x)^{n-1}-1\\ &=x\left((1+x)^n-1\right)+n(x^2+2x+n-1)(1+x)^{n-1}-x-1\\ f(x)&=x^2\left((1+x)^n-1\right)+nx(x^2+2x+n-1)(1+x)^{n-1}-x^2-x\\ f(1)&=2^n-1+n(n+2)2^{n-1}-2=(n^2+2n+2)2^{n-1}-3\\ &=\sum_{k=1}^{n} {(k+1)^2}\binom {n}{k}x^{k+1} \end{align}