`nC_1+nC_2+nC_3+n_4+.............+` equals

nC_(1)+nC_(2)+nC_(3)+............... equals to:

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)

if nC_12 = nC_8, then n is equal to

Find n if "^nC_2 = ^nC_3 .

If ^nC_12=^nC_8 , then n is equal to:

[(nC_(0)+nC_(3)+nC_(6)+.......)-(1)/(2)(nC_(1)+nC_(2)+nC_(4)+nC_ (5)+......)]^(2)+(3)/(4)(nC_(1)-nC_(2)+nC_(4)-nC_(5)+..... .)^(2)

If , ^nC_1,^nC_2 and ^nC_3 are in AP. Find the value of n.

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