$\sum_{r = 0}^{n} r^{2} .^{n} C_{r} p^{r} q^{n - r}, w h e r e p + q = 1,$ is simplified to:

n p q + n^{2} p^{2}

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n^{2} p^{2} q^{2} + n p

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n p (p + q)

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\frac{p (q + 1)}{2}

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Solution

The correct option is A $n p q + n^{2} p^{2}$
Given: $\sum_{r = 0}^{n} r^{2} x_{r} p^{r} q^{n - r}$

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

$= \sum_{r = 0}^{n} \frac{n! \times r \times r}{(r - 1)! (n - r)! r} p^{r} q^{n - r} + n^{2} p^{n}$

$= \sum_{r = 0}^{n} \frac{n (n - 1)!}{(r - 1)! (n - r)!} p \times (r p^{r - 1}) q \times q^{n - r - 1} + n^{2} p^{n}$

$= n p q \sum_{r = 0}^{n}^{n - 1} C_{r - 1} p \times (r p^{r - 1}) q^{n - r - 1} + n^{2} p^{n}$

$= n p q (^{n - 1} C_{0} p^{0} + q^{n - 2}) + (^{n - 2} C_{1} 2 p^{1} + q^{n - 3}) + (^{n - 1} C_{n - 1} p^{n - 2} q^{0} + n^{2} p^{n})$

$= n p q + n^{2} p^{n}$

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Similar questions

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