If p+q=1, then show that `sum_(r=0)^nr^2^nC_rp^rq^(n-r)=npq+n^2p^2`

Explanation

`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`