TITLE: Generating function of binomial coefficients ${n\choose5}$
QUESTION [5 upvotes]: 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.

REPLY [6 votes]: 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 $x=1/2$.