#11
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Re: Discovered series result
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. |
#12
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Re: Discovered series result
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 ? |
#13
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Re: Discovered series result
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 |
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