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

If p+q=1 then show that sum_(r=0)^(n)r^(2)C_(r)p^(r)q^(n-r)=npq+n^(2)p^(2)

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

If p+q=1, then value of sum_(r=0)^(n)r^(2)C(n,r)p^(r)q^(n-r) is (1)npq(2)np(1+q) (3) n^(2)p^(2)+npq(4)np^(2)+npq

If ""^(n)C_(0), ""^(n)C_(1),..., ""^(n)C_(n) denote the binomial coefficients in the expansion of (1 + x)^(n) and p + q = 1 , then sum_(r=0)^(n) ""r.^(n)C_(r) p^(r) q^(n-r) =

If ""(n)C_(0), ""(n)C_(1), ""(n)C_(2), ...., ""(n)C_(n), denote the binomial coefficients in the expansion of (1 + x)^(n) and p + q =1 sum_(r=0)^(n) r^(2 " "^n)C_(r) p^(r) q^(n-r) = .

Prove that sum_(r=0)^(n)r(n-r)C_(r)^(2)=n^(2)(^(2n-2)C_(n))

If (1+x)^n=sum_(r=0)^n C_rx^r then prove that sum_(r=0)^n (C_r)/((r+1)2^(r+1))=(3^(n+1)-2^(n+1))/((n+1)2^(n+1))

The sum of the series sum_(r=0) ^(n) ""^(2n)C_(r), is

Evaluate : sum_(r = 1)^(n) ""^(n)C_(r) 2^r