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^2p^2+npq` (4) `np^2+npq`.

The value of sum_(r=1)^n (^nP_r)/(r!) is

The value of sum_(r=1)^(n)(nP_(r))/(r!)=

The value of sum_(r=1)^(n)(nP_(r))/(r!) is

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

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

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)

The value of sum_(r=1)^(n)(sum_(p=0)^(n)nC_(r)^(r)C_(p)2^(p)) is equal to

The value of sum_(r=1)^n(sum_(p=0)^(r-1) ^nC_r ^rC_p 2^p) is equal to (a) 4^(n)-3^(n)+1 (b) 4^(n)-3^(n)-1 (c) 4^(n)-3^(n)+2 (d) 4^(n)-3^(n)