$$\text{Find} \ \ \sum_{k=0}^{n} (-1)^k k \binom{n}{k}^2$$
I expanded the binomial coefficients within the sum and got $$\binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + \dots + \binom{n}{n}^2$$
What does this equal to? I think this can help me evaluate the original sum.
Best Answer
First, use $k\binom{n\vphantom{1}}{k}=n\binom{n-1}{k-1}=n\binom{n-1}{n-k}$ $$ \sum_{k=0}^n(-1)^kk\binom{n}{k}^2 =n\sum_{k=0}^n(-1)^k\binom{n}{k}\binom{n-1}{n-k}\tag{1} $$ Next compute a generating function. The sum we want is the coefficient of $x^n$ $$ \begin{align} n\sum_{m,k}(-1)^k\binom{n}{k}\binom{n-1}{m-k}x^m &=n\sum_{m,k}(-1)^k\binom{n}{k}\binom{n-1}{m-k}x^{m-k}x^k\\ &=n\sum_k(-1)^k\binom{n}{k}(1+x)^{n-1}x^k\\ &=n(1+x)^{n-1}(1-x)^n\\ &=n\left(1-x^2\right)^{n-1}(1-x)\tag{2} \end{align} $$ The sum we want is the coefficient of $x^n$ in $(2)$: $$ \begin{align} \sum_{k=0}^n(-1)^kk\binom{n}{k}^2 &=\left\{\begin{array}{} n\binom{n-1}{n/2}(-1)^{n/2}&\quad\text{if $n$ is even}\\[6pt] n\binom{n-1}{(n-1)/2}(-1)^{(n+1)/2}&\quad\text{if $n$ is odd} \end{array}\right.\\[6pt] &=n\binom{n-1}{\lfloor n/2\rfloor}(-1)^{\lceil n/2\rceil}\tag{3} \end{align} $$