Find a closed formula to the sum: $\sum_{k=2}^{50}k (k-1){50 \choose k}$

Question

please help me with the formula, tried everything I know and I'm stuck!

$$\sum_{k=2}^{50}k(k-1){50 \choose k}$$

Thanks a lot!

Answer 1

There's a nice trick to finding closed forms like this. Consider $(1+x)^{50} = \displaystyle\sum_{k=0}^{50}{50\choose k}x^k$ by the binomial theorem. If we take the first derivative of both sides, we get $50(1+x)^{49} = \displaystyle\sum_{k=1}^{50}{50\choose k}kx^{k-1}$. Repeat to get $50(49)(1+x)^{48} = \displaystyle\sum_{k=2}^{50}{50\choose k}k(k-1)x^{k-2}$. Evaluating at $x=1$ should get you your answer: $50(49)(2^{48})$.

Answer 2

Using the identity $\binom{n}{k}=\frac nk\binom{n-1}{k-1}$, we get $$ \begin{align} \sum_{k=2}^{50}k(k-1)\binom{50}{k} &=\sum_{k=2}^{50}k(k-1)\binom{50}{k}\\ &=\sum_{k=2}^{50}50(k-1)\binom{49}{k-1}\\ &=\sum_{k=2}^{50}50\cdot49\binom{48}{k-2}\\[6pt] &=50\cdot49\cdot2^{48} \end{align} $$