$\sum_{r=0}^{19} (r+1)^4 \binom{20}{r+1}=\lambda. \, 2^{16}$ 
If $$\sum_{r=0}^{19} (r+1)^4 \binom{20}{r+1}=\lambda. \, 2^{16}$$ Find $\lambda$.

I wrote it as $20\sum_{r=0}^{19} (r+1)^3\binom{19}{r+1}$. Then expanding cube will do the work, but that will be really tedious. Any short way?
 A: Hint:
Set $r+1=n$
$$\sum_{n=1}^{20}n^3\binom{19}n$$
Now write $n^3=n(n-1)(n-2)+An(n-1)+Bn$  so that $$n^3\binom{19}n=n(n-1)(n-2)\binom{19}n+An(n-1)\binom{19}n+Bn\binom{19}n$$
$n=1\implies B=1$
$n=2\implies2^3=2A+2B\iff B=?$
A: It's not a very fast solution (not sure if one exists). If you consider
$$
G(x) = \sum_{k=1}^{19}\binom{19}{k}x^k = (1+x)^{19}
$$
then taking the derivative 3 times amounts to
$$
G'''(x) = \sum_{k=1}^{19}\binom{19}{k}k(k-1)(k-2)x^{k-3} \ \ \ \ \ \ (1)\\
G'''(x) = 19 \cdot 18 \cdot 17 (1+x)^{16} \ \ \ \ \ \ (2)
$$
Setting $x=1$ in(2) would have given you the solution for (1). But in your case you first need to expand
$$
k(k-1)(k-2) = k^3 -3k^2 +2k
$$
The expression with the first term is the one you actually have. The third one is easy, since by setting $x=1$ you get summand of the form $\binom{19}{k}k$. For the second though, you need to use the derivative again:
$$
\binom{19}{k}k^2 x^{k-3} = \frac{1}{x} \binom{19}{k}k^2x^{k-2} = \frac{1}{x^2}\frac{\partial}{\partial x}\binom{n}{k}k x^{k-3}
$$
After this you can sum the expression on $k$ and then take the partial derivative. The interchange of the sum and partial derivative is possible because the sum has a  finite upper bound and hence converges $\forall x \in \mathbb{R}$.
A: $$S=\sum_{r=0}^{19} (r+1)^4 {20 \choose r}=\sum_{k=1}^{20} k^4 {20 \choose k}$$
Let $$S'=\sum_{k=0}^{n} k^4 {n \choose k}$$
By binomial theorem $$(1+x)^n=\sum_{k=0}^{n} {n \choose k} x^k$$
Differentiate w.t. $x$ on both sides and multiply by
$$nx(1+x)^{n-1}= \sum_{k=0}^{n} k {n \choose k} x^{k}.$$
D.w.t.$x$, and multiply by $x$, to get
$$nx(1+x)^{n-1}+n(n-1)x^2(1+x)^{n-2} =\sum_{k=0}^{n} k^2 {n \choose k} x^k$$
Next, D.w.r.t $x$, and multiply by $x$. Further, again  D.w.r.t. $x$ and multiply by x to get
$$\sum_{k=0}^{n} k^4 {n \choose k}$$ as a function of $x$, finally put $x=1$ and $n=20$ to get the final answer.
