`2*^nC_0+2^2*(.^nC_1)/2+2^3*(.^nC_2)/3+...+2^(n+1)*(.^nC_n)/(n+1)=`

If (nC_0)/(2^n)+2.(nC_1)/2^n+3.(nC_2)/2^n+....(n+1)(nC_n)/2^n=16 then the value of 'n' is

Prove that ""^nC_0+2*""^nC_1+3*""^nC_2+...+(n+1)""^nC_n=(n+2)2^(n-1)

Prove that (""^nC_1)/(""^nC_0)+2*(""^nC_2)/(""^nC_1)+3*(""^nC_3)/(""^nC_2)+...+n*(""^nC_n)/(""^nC_(n-1))=frac{n(n+1)}{2}

nC_(0)-(1)/(2)(^(^^)nC_(1))+(1)/(3)(^(^^)nC_(2))-....+(- 1)^(n)(nC_(n))/(n+1)=

If s_n=""^nC_0+2*""^nC_1+3*""^nC_2+...+(n+1)*""^nC_n then find sum_(n=1)^oos_n .

(nC_(0))^(2)-(nC_(1))^(2)+(nC_(2))^(2)+....+(-1)^(n)(nC_(n))^(2)

If S_n=^nC_0.^nC_1+^nC_1.^nC_2+.....+^nC_(n-1).^nC_n and if S_(n+1)/S_n=15/4 , then the sum of all possible values of n is (A) 2 (B) 4 (C) 6 (D) 8

Prove that ""^nC_0+3*""^nC_1+5*""^nC_2+...+(2n+1)""^nC_n=(n+1)2^(n)

Show that : ""^nC_0*m-""^nC_1*(m-1)+""^nC_2*(m-2)-...+(-1)^n*""^nC_n*(m-n)=0

4^(n)C_(0)+4^(2)(nC_(1))/(3)+4^(3)(nC_(2))/(3)+4^(4)(nC_(3))/(4)+....+4^(n+1)(nC_(n))/(n+1) is equal to