Binomial Theorem By finding appropriate values for $x$ and $y$, evaluate
$$\sum_{k=0}^{n} k(k-1)\binom{n}{k}$$
I thought to take the derivative of $(1 + x)^n$ twice, but I noticed the index on $k$ remained at $0$. 
 A: The index on $k$ is not a problem. Look at, for example, the case $n=4$. $$(1+x)^4=1+4x+6x^2+4x^3+x^4$$ Differentiating twice, $$12(1+x)^2=(2)(1)(6)+(3)(2)(4)x+(4)(3)x^2$$ Note that the $k(k-1)$ is zero for $k=0$ and for $k=1$, so you can begin the sum at zero or at two, and it makes no difference. 
A: You have $$\sum_{k=0}^n {n\choose k} x^k = (1 + x)^n.$$
Differentiate. The constant term drops off, and you have
$$\sum_{k=1}^n {n\choose k} kx^{k-1} = n(1 + x)^{n-1}.$$
Do it again to get
$$\sum_{k=2}^n {n\choose k} x^{k-2} = n(n-1)(1 + x)^{n-2}.$$
You can take $x=0$, exhibit a little care, and see all.
A: That works fine. 
$$\begin{align*}
\frac{d^2}{dx^2}\left((1+x)^n\right)&=\frac{d^2}{dx^2}\left(\sum_{k\ge 0}\binom{n}kx^k\right)\\\\
&=\frac{d}{dx}\left(\sum_{k\ge 0}k\binom{n}kx^{k-1}\right)\\\\
&=\sum_{k\ge 0}k(k-1)\binom{n}kx^{k-2}\;.
\end{align*}$$
(Recall that $\binom{n}k=0$ if $k>n$, so there’s no need to worry about the upper limit.)
Now substitute $x=1$ to get
$$n(n-1)2^{n-2}=\sum_{k\ge 0}k(k-1)\binom{n}k\;.$$
It’s perfectly true that the first two terms are $0$, and we could just as well write
$$\sum_{k=2}^nk(k-1)\binom{n}k\;,$$
but that’s unnecessarily ugly. It’s also true that if we were differentiating by hand a small example, we’d never even write down those terms: we don’t normally write
$$(-1)(0)x^{-2}+(0)(1)4x^{-1}+(1)(2)6x^0+(2)(3)4x^1+(3)(4)x^2$$
for the second derivative of $1+4x+6x^2+4x^3+x^4$, but those factors of $0$ ensure that there’s no actual harm in it.
A: Without using derivatives,
$$\text{ for }k\ge 2, k(k-1)\binom nk=\frac{k(k-1)n!}{(n-k)!k!}=\frac{n(n-1)k(k-1)(n-2)!}{\{n-2-(k-2)\}!(k-2)!k(k-1)}=n(n-1)\binom{n-2}{k-2}$$
So, $$\sum_{0\le k\le n}k(k-1)\binom nk$$
$$=\sum_{2\le k\le n}k(k-1)\binom nk$$
$$=\sum_{2\le k\le n}n(n-1)\binom{n-2}{k-2}$$
$$=n(n-1)\sum_{0\le r\le n-2}\binom {n-2}r$$
$$=n(n-1)(1+1)^{n-2}$$
