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

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 $$n^3\binom{19}n=n(n-1)(n-2)\binom{19}n+An(n-1)\binom{19}n+Bn\binom{19}n$$
$n=1\implies B=1$
$n=2\implies2^3=2A+2B\iff B=?$
