Distribution of a binomial variable squared

Usually we derive the variance of the binomial distribution from the calculation of its second moment, so to refer to the variance in order to get the second moment would be somewhat circular reasoning.

The direct calculation is as follows. Consider $$\begin{align*} {\rm E}[X(X-1)] &= \sum_{k=0}^n k(k-1) \binom{n}{k} p^k (1-p)^{n-k} \\ &= \sum_{k=2}^n \frac{n(n-1)(n-2)!}{(k-2)!((n-2)-(k-2))!} p^2 p^{k-2} (1-p)^{(n-2)-(k-2)} \\ &= n(n-1)p^2 \sum_{k'=0}^{n-2} \frac{(n-2)!}{(k')!((n-2)-k')!} p^{k'} (1-p)^{(n-2)-k'} \\ &= n(n-1)p^2 \sum_{k'=0}^{n-2} \binom{n-2}{k'} p^{k'} (1-p)^{(n-2)-k'} \\ &= n(n-1)p^2, \end{align*}$$ because the summand in the penultimate step is simply the probability mass function of a binomial random variable with parameters $n-2$ and $p$, so it sums to $1$ (which is also evident via the binomial theorem). Indeed, this calculation is easily generalized: $${\rm E}[X(X-1)\cdots(X-m)] = \frac{n!}{(n-m-1)!}p^{m+1},$$ for which the above is the special case $m = 1$. For $m = 0$, we easily get ${\rm E}[X] = np$. Then combining these results via the linearity of expectation gives $${\rm E}[X^2] = {\rm E}[X(X-1) + X] = {\rm E}[X(X-1)] + {\rm E}[X] = n(n-1)p^2 + np.$$

Hint: You can use the fact that $$ Var(X)=E(X^{2})-E(X)^{2} $$

to find $E(X^{2})$. This is assuming you know both the mean and variance of a binomial random variable

$\begin{align} \text{E}(X^2) & = \sum\limits_{k = 0}^n k^2P[X = k]\\ & = \sum\limits_{k = 0}^n k^2\,^nC_kp^kq^{n - k}\\ & = \sum\limits_{k = 1}^n k^2\, \dfrac{n!}{k! (n - k)!} p^kq^{n - k}\\ & = \sum\limits_{k = 1}^n k \dfrac{n!}{(k - 1)! (n - k)!} p^kq^{n - k}\\ & = \sum\limits_{k = 1}^n (k - 1 + 1) \dfrac{n!}{(k - 1)! (n - k)!} p^kq^{n - k}\\ & = \sum\limits_{k = 1}^n (k - 1) \dfrac{n!}{(k - 1)! (n - k)!} p^kq^{n - k} + \sum\limits_{k = 1}^n \dfrac{n!}{(k - 1)! (n - k)!} p^kq^{n - k}\\ & = \sum\limits_{k = 2}^n \dfrac{n!}{(k - 2)! (n - k)!} p^kq^{n - k} + \sum\limits_{k = 1}^n \dfrac{n!}{(k - 1)! (n - k)!} p^kq^{n - k}\\ & = n(n-1)p^2\sum\limits_{k = 2}^n \dfrac{(n-2)!}{(k - 2)! (n - k)!} p^{k-2}q^{(n-2) - (k-2)} +\\& \qquad np\sum\limits_{k = 1}^n \dfrac{(n-1)!}{(k - 1)! (n - k)!} p^{k-1}q^{(n-1) - (k-1)}\\ & = n(n-1)p^2\sum\limits_{k=2}^n \, ^{n-2}C_{k-2}p^{k-2}q^{(n-2) - (k-2)} + np\sum\limits_{k = 1}^n \, ^{n-1}C_{k-1} p^{k-1}q^{(n-1) - (k-1)}\\ & = n(n - 1)p^2(p + q)^{n - 2} + np(p + q)^{n - 1}\\ & = n^2p^2 - np^2 + np\\ & = \boxed{n^2p^2 + npq} \end{align}$