Finding closed form of $\sum\limits_{r=0}^n r^2 \binom nr p^{n-r}q^r$ where $p>0$ and $q=1-p$

Question

Here is a question from Indian Statistical Institute (ISI) Entrance Exam CSA-2020, Question 11:

$\binom n0, \binom n1,..., \binom nn$ denote the binomial coefficients in the expansion of $(1 + x)^n \ \text{where} \ p > 0$ is a real number and $q = 1 − p$ then $$ \sum\limits_{r=0}^n r^2 \binom nr p^{n-r}q^r $$ is equal to

$ (A)\ np^2q^2 \ (B)\ n^2p^2q^2 \ (C)\ npq + n^2p^2 \ (D)\ npq + n^2q^2 $

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My Thoughts: (1)

If I put the formula for $\binom nr$ in the given sum. I am getting

$$ \sum\limits_{r=0}^n r^2 \frac{n!}{r!(n-r)!} p^{n-r}q^r $$

How do I proceed after this?

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Edit:

Taking help from here and the comments below I wrote the following, although I am stuck in the last step, please help me out.

$$ (1+x)^n=\sum\limits_{r=0}^n \binom nr x^r $$ Taking derivative on both sides we have: $$ n(1+x)^{n-1}=\sum\limits_{r=1}^n r\binom nr x^{r-1} $$ Multiplying both sides by $x$ we get: $$ nx(1+x)^{n-1}=\sum\limits_{r=1}^n r \binom nr x^r $$ Taking derivative another time: $$ n(1+x)^{n-1}+nx(n-1)(1+x)^{n-2}=\sum\limits_{r=1}^n r^2 \binom nr x^{r-1} $$ Multiplying both sides by $x$ we get: $$ nx(1+x)^{n-1}+nx^2(n-1)(1+x)^{n-2}=\sum\limits_{r=1}^n r^2 \binom nr x^r $$

Given that $p>0$ and $q=1-p$ . So let $x=\frac qp$ then we have: $$ n\frac qp\left(1+\frac qp\right)^{n-1}+ n\left(\frac qp\right)^2(n-1)\left(1+\frac qp\right)^{n-2} = \sum\limits_{r=1}^n r^2 \binom nr \left(\frac qp \right)^r $$

$$ \implies n\frac qp\left(\frac{p+q}{p}\right)^{n-1}+ n\left(\frac qp\right)^2(n-1)\left(\frac{p+q}{p}\right)^{n-2} = \sum\limits_{r=1}^n r^2 \binom nr \left(\frac qp \right)^r $$

$$ \implies n\frac qp \left(\frac{1}{p}\right)^{n-1}+ n\left(\frac qp\right)^2(n-1)\left(\frac{1}{p}\right)^{n-2} = \sum\limits_{r=1}^n r^2 \binom nr \left(\frac qp \right)^r \ \ \ \ \ [\text{Since}\ p+q=1] $$

$$ \implies n\frac{q}{p^n}+ n(n-1)\frac{q^2}{p^n} = \sum\limits_{r=1}^n r^2 \binom nr \left(\frac qp \right)^r $$

$$ \implies nq+ n(n-1)q^2 = p^n\sum\limits_{r=1}^n r^2 \binom nr \left(\frac qp \right)^r $$

$$ \implies nq+ n(n-1)q^2 = \sum\limits_{r=1}^n r^2 \binom nr p^{n-r}q^r $$

$$ \implies n(q-q^2)+ n^2q^2 = \sum\limits_{r=1}^n r^2 \binom nr p^{n-r}q^r $$

$$ \implies n\{(1-p)-(1-p)^2\}+ n^2q^2 = \sum\limits_{r=1}^n r^2 \binom nr p^{n-r}q^r $$

$$ \implies n\{(1-p)-(1+p^2-2p)\}+ n^2q^2 = \sum\limits_{r=1}^n r^2 \binom nr p^{n-r}q^r $$

$$ \implies np(1-p)+ n^2q^2 = \sum\limits_{r=1}^n r^2 \binom nr p^{n-r}q^r $$

$$ \implies npq+ n^2q^2 = \sum\limits_{r=1}^n r^2 \binom nr p^{n-r}q^r $$

Hence option (D) is correct.

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This question is almost similar to this question but is slightly different. So I have added this question.

Answer 1

Hint 1: $$r\binom{n}{r}=n\binom{n-1}{r-1}$$

Hint 2: $$r\binom{n-1}{r-1}=(r-1+1)\binom{n-1}{r-1}=\text{??}$$

Also note that you can take some $p$s and $q$s out of the summation (since they're constants) to match the upper index of the binomial.

Answer 2

You probably know that the expected number of successes in $n$ independent trials with probability $p$ of success on each trial is $np,$ and that the variance of the number of successes is $npq=np(1-p).$

Now recall that the expected value of the square of a random variable is the variance plus the square of the expected value.

Therefore you get $npq + n^2p^2.$

Answer 3

Let $$f(q) = (p+q)^n = \sum_{r = 0}^{n} \binom{n}{r}p^{n-r}q^r$$

Then $$q \frac{df}{dq} = q\sum_{r = 0}^{n} r\binom{n}{r}p^{n-r}q^{r-1} = \sum_{r = 0}^{n} r\binom{n}{r}p^{n-r}q^{r}$$ What just happened, by applying $q\frac{d}{dq}$ to $f(q)$, is each power $r$ of $q$ has been brought down from the exponent to multiply the coefficient of $q^r$. By doing this twice, one gets $$q\frac{d}{dq}\left(q\frac{df}{dq}\right) = q\left(\frac{df}{dq}+q\frac{d^2 f}{dq^2}\right) = q\frac{df}{dq}+q^2\frac{d^{2} f}{dq^2} = \sum_{r = 0}^{n} r^2 \binom{n}{r}p^{n-r}q^r$$ Plug in the original definition for $f(q)$ into this equation and evaluate at $q = 1-p$, then in your multiple choices, plug in $q = 1-p$, and then compare.