Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $$k=5,\qquad t=\frac12.$$

share|cite|improve this answer
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 $x=1/2$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.