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If p+q=1, then show that sum(r=0)^n r^2^...

If `p+q=1,` then show that `sum_(r=0)^n r^2^n C_rp^r q^(n-r)=n p q+n^2p^2dot`

Text Solution

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`underset(r=0)overset(n)sumr^(2).^(n)C_(r)p^(r)q^(n-r)`
`= underset(r=0)overset(n)sumnr.^(n-1)C_(r-1)p^(r)q^(n-r)`
`= n underset(r=0)overset(n)sum[(r-1)+1)]^(n-1)C_(r-1)p^(r)q^(n-r)`
`= n underset(r=0)overset(n)sum[(r-1)^(n-1)C_(r-1)+.^(n+1)C_(r-1)]p^(r)q^(n-r)`
`= n underset(r=0)overset(n)sum[(n-1).^(n-2)C_(r-1)p^(r-2)q^(n-r)+np underset(r=0)overset(n)sum.^(n-1)C_(r-1)p^(r-1)]q^(n-r)`
`= p^(2)n(n-1)(p+q)^(n-2)+np(p+q)^(n-1)`
`=p^(2)n(n-1)+np`
`=p^(2)n^(2)+np(1-p)`
`= p^(2)n^(2)+npq`
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