P and Q are integers

Method 1

$\displaystyle \begin{aligned}\sum\limits_{r=0}^nr^2\binom{n}{r}p^rq^{n-r} &=\sum\limits_{r=0}^nnp\binom{n-1}{r-1}rp^{r-1}q^{n-r}\\ &=np\sum\limits_{r=0}^n\dfrac{\mathrm{d}}{\mathrm{d}p}\left(\binom{n-1}{r-1}p^rq^{n-r}\right)\\ &=np\dfrac{\mathrm{d}}{\mathrm{d}p}\left[p(p+q)^{n-1}\right]\\ &=np\left[(p+q)^{n-1} + (n-1)p(p+q)^{n-2}\right]\\ &=np + n(n-1)p^2\end{aligned}\\$

$\Huge\displaystyle \therefore\boxed{\sum\limits_{r=0}^nr^2\binom{n}{r}p^rq^{n-r} = np + n(n-1)p^2}$

If the random variable $X$ has the Binomial distribution $B(n,p)$ , we are asked to calculate $E[X^2] = \mathrm{Var}[X] + E[X]^2$ . It is standard bookwork that $X$ has mean $np$ and variance $npq$ , and so our answer is $npq + n^2p^2$ . For the given values of $n,p$ , this makes the answer $\tfrac{2057}{125} = \boxed{16.456}$ .

To show that the binomial distribution has the stated mean and variance, note that if $X \sim B(n,p)$ , then $X = Z_1 + Z_2 + \cdots + Z_n$ , where $Z_j$ is equal to $1$ if the $j$ th trial is a success, and $0$ otherwise. Then $Z_1,Z_2,\ldots,Z_n$ are independent and identically distributed, and $E[Z_1^2] = E[Z_1] = P[Z_1 = 1] = p$ so that $E[Z_1] \; = \; p \hspace{2cm} \mathrm{Var}[Z_1] \; = \; p - p^2 \; = \; pq$ Thus $\begin{aligned} E[X] & = \; \sum_{j=1}^n E[Z_j] \; = \; np \\ \mathrm{Var}[X] &= \; \sum_{j=1}^n \mathrm{Var}[Z_j] \; = \; npq \end{aligned}$