How can I prove this property of binomial coefficients?

Question

I was playing around with binomial coefficients and binomial expansions and I came across an interesting identity:

$$ \sum_{k=0}^n\frac{1}{k+1}{n \choose k}x^k=\frac{(x+1)^{n+1}-1}{(n+1)x}$$

I have verified this for a few different values of n and wolfram alpha seems to agree (see here) but I'm not quite sure how to go about proving it.

I've tried writing the binomial coefficient out in its factorial definition and manipulating the factorials from there but it never ended up getting me anywhere — ASKASK, Apr 08 '15 at 15:02
Probably by playing around with derivatives you can make that form. — FreeMind, Apr 08 '15 at 15:02

Hypergeometricx · Accepted Answer · 2015-04-08T15:12:38.790

3

$$\begin{align} (x+1)^{n+1}-1 &=\sum_{k=0}^{n+1}{n+1\choose k}x^k-1\\ &=\sum_{k=1}^{n+1}{n+1\choose k}x^k\\ &=\sum_{k=0}^{n}{n+1\choose k+1}x^{k+1}\\ &=\sum_{k=0}^{n}\frac{n+1}{k+1}{n\choose k}x^{k+1}\\ &=(n+1)x\sum_{k=0}^{n}\frac 1{k+1}{n\choose k}x^k\\ \sum_{k=0}^{n}\frac 1{k+1}{n\choose k}x^k &=\frac{(x+1)^{n+1}-1}{(n+1)x}\qquad\blacksquare \end{align}$$

edited Apr 08 '15 at 15:12

answered Apr 08 '15 at 15:11

Hypergeometricx

22,657

Wow. Vastly simple. Thanks! – ASKASK Apr 08 '15 at 15:11
Just one slight thing, shouldn't the upper limit of the summation change to $n$ on the third line? – ASKASK Apr 08 '15 at 15:12
@ASKASK Thank you! Yes, edited. – Hypergeometricx Apr 08 '15 at 15:13

score 1 · Answer 2 · answered Apr 08 '15 at 16:32

Define $$f(x)= \sum_{k=0}^n\frac{1}{k+1}{n \choose k}x^k$$

$$g(x) = xf(x) = \sum_{k=0}^n\frac{1}{k+1}{n \choose k}x^{k+1}$$

Then we have $g'(x) =\sum_{k=0}^n{n \choose k}x^{k} = (1+x)^n$.

Since $g(0) = 0$, we get $g(x) = \frac{(1+x)^{n+1} -1}{n+1}$, then the expression of $f(x)$ accordingly.

How can I prove this property of binomial coefficients?

2 Answers2

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