Discovered series result

[T50_Omaha8]

I may be a dunce but I can't get past

((x^k)/(1-x))*(C(k,0)x^0+C(k,1)x^1+...+C(k,k)x^(k))

which I only got using combinatorial identities and geometric sums. Looks like it's taking shape though.

[jay_shark]

Ok I figured out another way to solve this .

The expression is equivalent to

{sum n=k to infinity}C(n,k)x^(n-k)*(1-x)^(k+1)=1

Suppose you have a bias coin with probability x of landing heads and probability (1-x) of landing tails . You keep on flipping this coin infinitely many times until you've flipped k+1 tails . Certainly the probability of this happening is 1 which is the rhs of the equation . Or equivalently ,
you may hit k+1 tails on your (k+1)st turn ,
k+1 tails on your (k+2)nd turn
k+1 tails on your (k+3)rd turn

This is exactly the lhs of the equation .

Why didn't I think of this before ?

[jason1990]

Oh, yes. Nice observation: negative binomial.

{sum n=k to infinity}C(n,k)x^(n-k)*(1 - x)^(k+1)
= {sum n=0 to infinity}C(n+k,k)x^n*(1 - x)^(k+1)
= {sum n=0 to infinity}((n + k)!/(n!k!))x^n*(1 - x)^(k+1)
= {sum n=0 to infinity}((n + k)(n + k - 1)...(k + 1)/n!)x^n*(1 - x)^(k+1)
= (1 - x)^(k+1){sum n=0 to infinity}((-k - 1)(-k - 2)...(-k - n))/n!)(-x)^n
= (1 - x)^(k+1){sum n=0 to infinity}C(-k-1,n)(-x)^n
= (1 - x)^{k+1}(1 + (-x))^{-k-1}
= 1