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How to prove easily this identity for (almost classical) series with binomial coefficients:

$$ \sum_{n=5}^\infty \dfrac{\binom{n}{5}}{2^{n+1}} = 1 . $$

Thank you. Any smart proof would be much appreciated.

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up vote 11 down vote accepted

Differentiating $k$ times the identity, valid for $|t|\lt1$, $$\sum_{n\geqslant0}t^n=\frac1{1-t}$$ yields $$ \sum_{n\geqslant k}{n\choose k}t^{n+1}=\frac{t^{k+1}}{(1-t)^{k+1}}=\left(\frac{t}{1-t}\right)^{k+1}.$$ Use this for $$k=5,\qquad t=\frac12.$$

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Yes, thank you. Nice. Hard to believe that shorter proof exists. – Oleg567 Jun 13 '14 at 9:17

The Binomial Theorem says $$(1-x)^{-6}=\sum_0^{\infty}{n+5\choose5}x^n$$ Rewrite as $$(1-x)^{-6}=\sum_{n=5}^{\infty}{n\choose5}x^{n-5}$$ Multiply by $x^6$ to get $$x^6(1-x)^{-6}=\sum_{n=5}^{\infty}{n\choose5}x^{n+1}$$ Now let $x=1/2$.

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Thank you for referring to Binomial Theorem. Nice too. I'll bookmark "Binomial Theorem" from Wiki now. I didn't know yet about series like first one. – Oleg567 Jun 13 '14 at 10:58

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