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Question: If \(p+q = 1\), then \(\sum_{r=0}^{n} r^3 \binom{n}{r} p^r q^{n-r} =\)...

If p+q=1p+q = 1, then ∑r=0nr3(nr)prqn−r=\sum_{r=0}^{n} r^3 \binom{n}{r} p^r q^{n-r} =

A

np(n2p+3(n−1)p+1)n p \bigl(n^2 p + 3(n-1)p + 1\bigr)

B

np((n2−n)p2+2(n−1)p+1)n p\bigl((n^2 - n) p^2 + 2(n-1)p + 1\bigr)

C

np((n2−3n+2)p2+2(n−1)p+1)n p\bigl((n^2 - 3n + 2) p^2 + 2(n-1)p + 1\bigr)

D

np((n2−3n+2)p2+3(n−1)p+1)n p\bigl((n^2 - 3n + 2) p^2 + 3(n-1)p + 1\bigr)

Answer

np((n2−3n+2)p2+3(n−1)p+1)n p\bigl((n^2 - 3n + 2) p^2 + 3(n-1)p + 1\bigr)

Explanation

Solution

Using the third moment of a binomial distribution
For X∼Binomial(n,p)X\sim \mathrm{Binomial}(n,p):

E[X3]=np(1−p)(1−2p)  +  3n2p2(1−p)  +  n3p3.E[X^3] = n p(1-p)(1-2p) \;+\; 3n^2 p^2(1-p)\;+\;n^3p^3.

Let q=1−pq=1-p. Then

E[X3]=np q (1−2p)  +  3n2p2 q  +  n3p3.E[X^3] = n p\,q\,(1-2p) \;+\;3n^2p^2\,q\;+\;n^3p^3.

Expand each term:

np q (1−2p)=np (1−p) (1−2p)=np (1−3p+2p2),n p\,q\,(1-2p) = n p\,(1-p)\,(1-2p) = n p\,(1 - 3p + 2p^2), 3n2p2 q=3n2p2(1−p)=3n2p2−3n2p3,3n^2p^2\,q = 3n^2p^2(1-p)=3n^2p^2 -3n^2p^3, n3p3=n3p3.n^3p^3 = n^3p^3.

Summing coefficients of like powers of pp:

E[X3]=np+(3n2−3n) p2+(n3−3n2+2n) p3=np[1+3(n−1)p+(n2−3n+2)p2].E[X^3] = n p + (3n^2 - 3n)\,p^2 + (n^3 - 3n^2 +2n)\,p^3 = n p\Bigl[1 + 3(n-1)p + (n^2-3n+2)p^2\Bigr].

Hence the required sum is

np((n2−3n+2)p2+3(n−1)p+1).n p\bigl((n^2 - 3n + 2)p^2 + 3(n-1)p + 1\bigr).