Prove that
`(1^(2))/(3).^(n)C_(1)+(1^(2) + 2^(2))/(7).^(n)C_(2)+(1^(2)+2^(2)+3^(2))/(7).^(n)C_(3)+"...."`
`+(1^(2)+2^(3)+"....."+n^(2))/(2n+1).^(n)C_(n) = (n(n+3))/(6)2^(n-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)S=underset(r=1)overset(n)sum(1^(2)+2^(2)+"...."+r^(3))/(2r+1).^(n)C_(r)`
`= underset(r=1)overset(n)sum(r(r+1)(2r+1))/(6(2r+1)).^(n)C_(r)`
`= 1/6underset(r=1)overset(n)sumr(r+1).^(n)C_(r)`
`= 1/6underset(r=1)overset(n)sum(r+1).n..^(n-1)C_(r-1)`
`=1/6n underset(r=1)overset(n)sum((r-1)+2)^(n-1)C_(r-1)`
`=1/6n.underset(r=1)overset(n)sum((r-1)..^(n-1)C_(r-1)+2..^(n-1)C_(r-1))`
`= 1/6n.underset(r=1)overset(n)sum((n-1)..^(n-2)C_(r-2)+2..^(n-1)C_(r-1))`
`=1/6n.(n-1).2^(n-2)+(n)/(3).2^(n-1)=1/6n(n+3)2^(n-2)`