`nC_1+nC_2+nC_3+.................`equals to:

nC_(1)+nC_(2)+nC_(3)+n_(4)+............+ equals

The sum to (n + 1) terms of the series 3.nC_0 - 8.nC_1 +13.nC_2 -18.nC_2 +........ is

If (1 + x)^n= ""^nC_0+""^nC_1x+""^nC_2x^2+.............+ ""^nC_n x^n , prove that : ""^nC_1-2""^nC_2+3""^nC_3-.......+(-1)^(n-1) n ""^nC_n=0 .

Let a=(4^(1/401)-1) and let b_n = nC_1 + nC_2 . a+ nC_3. a^2+......+ nC_n .a^(n-1). Find the value of (b_2006 – b_2005)

2nC_(n+1)+2*2nC_(n)+2nC_(n-1) is equals to:

2nC_(n+1)+2*2nC_(n)+2nC_(n-1) is equals to:

Find n if "^nC_2 = ^nC_3 .

If n is a positive integer, then ^nC_1+^nC_2+....+^nC_n is equal to

(n+2)nC_0(2^(n+1))-(n+1)nC_1(2^(n))+(n)nC_2(2^(n-1))-.... is equal to

nC_(0)-2.3nC_(1),+3.3^(2)nC_(2)-4.3^(3)nC_(3)+.........+(-1)(n+1)nC_(n)3^(n) is eqaul to.