Prove that $\sum_{k=0}^{n}\frac{\binom{n}{k}}{k+1}=\frac{2^{n+1}-1}{n+1}$

Question

I want to prove the following
$$\sum_{k=0}^{n}\frac{\binom{n}{k}}{k+1}=\frac{2^{n+1}-1}{n+1}$$ I need some hint how to start.
thanks!

You can use induction and binomial identities: http://en.wikipedia.org/wiki/Binomial_coefficient#Identities_involving_binomial_coefficients — ulead86, Aug 29 '13 at 07:32

score 5 · Accepted Answer · answered Aug 29 '13 at 20:18

Following your hint to introduce integrals we have $$\sum_{k=0}^n \frac{1}{k+1} \binom{n}{k} = \sum_{k=0}^n \binom{n}{k} \int_0^1 x^k dx$$ which is $$\int_0^1 \sum_{k=0}^n \binom{n}{k} x^k dx = \int_0^1 (x+1)^n dx = \left. \frac{(x+1)^{n+1}}{n+1} \right|_0^1 = \frac{2^{n+1}-1}{n+1}.$$

score 1 · Answer 2 · answered Aug 29 '13 at 07:33

1

HINT: Multiply both sides by $n+1$, and note that $$\frac{n+1}{k+1}\binom{n}k=\binom{n+1}{k+1}\;.$$

answered Aug 29 '13 at 07:33

Brian M. Scott

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$2^{n+1}-1 = \binom{n+1}{k+1}$? – Ofir Attia Aug 29 '13 at 07:33
@Ofir: No, certainly not. The lefthand side is sum, not a single term. – Brian M. Scott Aug 29 '13 at 07:34

Prove that $\sum_{k=0}^{n}\frac{\binom{n}{k}}{k+1}=\frac{2^{n+1}-1}{n+1}$

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