$$ \sum_{k=0}^{n-1}{n+k-1\choose k}\frac 1{2^{n+k}}=\frac12$$
To be honest, I can't really get started,
I would like to ask everyone to give me an idea of how to solve it, give me a starting push, thank you.
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Welcome to MSE. Please type your questions instead of posting images. Images can't be browsed and are not accessible to those using screen readers. If you need help formatting math on this site, here's a tutorial – saulspatz Nov 30 '20 at 00:08
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thank you so much, im a beginner:) – panda Nov 30 '20 at 00:10
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understandable format? – panda Nov 30 '20 at 00:13
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Did I guess the intended formulas right? – Hagen von Eitzen Nov 30 '20 at 00:14
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Induction maybe? Prove it for $n=1$, then use that result to prove that if it holds for $n$ it will also hold in the $n+1$ case. – user525966 Nov 30 '20 at 00:15
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@HagenvonEitzen yeees :)))) thank you very much for your help – panda Nov 30 '20 at 00:16
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@user525966 its okay, but induction was in my head too, it is hard for me to do in practice – panda Nov 30 '20 at 00:19
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@panda take out $\frac{1}{2^n}$ and look at the summation; this seems to have come from a stars and bars problem. – across Nov 30 '20 at 00:26
2 Answers
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By the way, the product being summed looks like the PMF of the negative binomial distribution, that to flip a fair coin repeatedly until having $n$ heads, there's $\frac 12$ probability that there were at most $n-1$ tails before the $n$th head.
Let $\Pr(A)$ be the probability that there were at most $n-1$ tails before the $n$th head.
Then $1-\Pr(A)$ would be the probability that the $n$th tail appears while there were at most $n-1$ heads.
For a fair coin, head and tail are symmetric, and so
$$\Pr(A) = 1-\Pr(A) = \frac 12$$
peterwhy
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1[+1] I like very much the insight given by your proof. Proofs by induction are given here but none of them, though interesting, is very simple. – Jean Marie Nov 30 '20 at 01:10
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we rewrite the sum like this ;$$\sum_{k=0}^{k=n-1}\binom{n+k-1}{n-1}\frac{1}{2^{n+k}}$$
this is just the coefficient of $x^{n-1}$ in expansion $$\frac{{(1+x)}^{n-1}}{2^n}+\frac{{(1+x)}^n}{2^{n+1}}...+\frac{{(1+x)}^{2n-2}}{2^{2n-1}}$$
Recognise this as a GP: hence we seek the coefficient of $x^{n-1}$ in $$\frac{1}{2^{n-1}}{(1+x)}^{n-1}\frac{1-{(\frac{x+1}{2})}^n}{1-x}$$
or $$\frac{{(1+x)}^{n-1}(1+x+x^2..)-{(1+x)}^{2n-1}(1+x+x^2+x^3....)}{2^{2n-1}}$$ The coefficent is $$\frac{\left(\binom{n-1}{0}+\binom{n-1}{1}..+\binom{n-1}{n-1}\right)-\left(\binom{2n-1}{0}+\binom{2n-1}{1}+...\binom{2n-1}{n-1}\right)}{2^{2n-1}}$$ $$=\frac{2^{2n-1}-\frac{1}{2}2^{2n-1}}{2^{2n-1}}=\frac{1}{2}$$
Albus Dumbledore
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