Expected product of head and tail counts

Toss a fair coin $n$ times; let $X$ be the number of heads and $Y$ the number of tails. Writing $\mathbb{E}[XY]=\dfrac{an^2+bn+c}{d}$ in lowest terms with integers $a,b,c,d$, find $a+b+c+d$.

$4$. Since $Y=n-X$ with $X\sim\text{Bin}(n,\tfrac12)$, we get $XY=nX-X^2$. Using $\mathbb{E}X=\tfrac n2$ and $\mathbb{E}X^2=\tfrac n4+\tfrac{n^2}{4}$ gives $\tfrac{n^2-n}{4}$, so $(a,b,c,d)=(1,-1,0,4)$ and the sum is $4$.