If $^{n} C_{0},^{n} C_{1},^{n} C_{2}, \dots,^{n} C_{n}$ denote the binomial coefficients in the expansion of $(1 + x)^{n}$ and $p + q = 1,$ then $n \sum r = 0 r^{2}^{n} C_{r} p^{r} q^{n - r}$ is

n p

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n p q

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

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None of these

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Solution

The correct option is C $n^{2} p^{2} + n p q$
We have $n \sum r = 0 r^{2}^{n} C_{r} p^{r} q^{n - r}$
$= n \sum r = 0 [r (r - 1) + r]^{n} C_{r} p^{r} q^{n - r}$
$= n \sum r = 0 r (r - 1)^{n} C_{r} p^{r} q^{n - r} + n \sum r = 0 r \cdot^{n} C_{r} p^{r} q^{n - r}$
$= n \sum r = 0 r (r - 1) \frac{n}{r} \cdot \frac{n - 1}{r - 1}^{n - 2} C_{r - 2} p^{r} q^{n - r} + n \sum r = 0 r \cdot \frac{n}{r}^{n - 1} C_{r - 1} p^{r} q^{n - r}$
$= n (n - 1) p^{2} (q + p)^{n - 2} + n p (q + p)^{n - 1}$
$= n (n - 1) p^{2} + n p [∵ q + p = 1]$
$= n^{2} p^{2} - n p^{2} + n p$
$= n^{2} p^{2} + n p q [∵ q + p = 1]$

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Similar questions

Q. If

^{n} C_{0},^{n} C_{1},^{n} C_{2}, \dots,^{n} C_{n}

denote the binomial coefficients in the expansion of

(1 + x)^{n}

and

p + q = 1

, then

n \sum r = 0 r^{2}^{n} C_{r} p^{r} q^{n - r}

Q. If

^{n} C_{0},^{n} C_{1},^{n} C_{2}, \dots,^{n} C_{n}

denote the binomial coeffients in the expansion of

(1 + x)^{n}

and

p + q = 1,

then

n \sum r = 0 r^{n} C_{r} p^{r} q^{n - r}

Find the coefficient of $x^{n - 2}$ in
$(^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} . . . . .^{n} C_{n} x^{n}) \times (^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} . . . . .^{n} C_{n} x^{n})$

Q. A variable takes the values 0, 1, 2, 3, ....n with frequencies proportional to the binomial coefficients

^{n} c_{0},^{n} c_{1},^{n} c_{2},^{n} c_{3}, . . . . . . .^{n} c_{n}

then variance

σ^{2} =

Q. IF

S_{n} =^{n} C_{0} .^{n} C_{1} +^{n} C_{1} .^{n} C_{2} + . . . . +^{n} C_{n - 1} .^{n} C_{n} a n d \frac{S_{n + 1}}{S_{n}} = \frac{15}{4}

, then n =

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If $^{n} C_{0},^{n} C_{1},^{n} C_{2}, \dots,^{n} C_{n}$ denote the binomial coefficients in the expansion of $(1 + x)^{n}$ and $p + q = 1,$ then $n \sum r = 0 r^{2}^{n} C_{r} p^{r} q^{n - r}$ is