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There is this identity

$$1 -\frac{1}{2}\binom{n}{1}+\frac{1}{3} \binom{n}{2}- \frac{1}{4}\binom{n}{3}+....+(-1)^n \frac{1}{n+1}\binom{n}{n}$$

And we are supposed to prove it using these two identities

$$k\binom{n}{k} = n\binom{n-1}{k-1}$$

and

$$\binom{n}{0} + \binom{n}{1} + \binom{n}{2} +....+ \binom{n}{n} = 2^n$$

I have been working on this problem for a long time. Can you guys help me?

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  • $\begingroup$ Your first expression? $\endgroup$ – user418131 Oct 3 '17 at 19:24
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    $\begingroup$ Is this correct as written? I do not see an equality to prove $\endgroup$ – TomGrubb Oct 3 '17 at 19:30
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\begin{eqnarray*} \frac{1}{i+1}= \int_0^1 x^i dx \end{eqnarray*} Sub this into the sum & interchange the order of the sum & the integral \begin{eqnarray*} \sum_{i=0}^{n} (-1)^i\binom{n}{i} \frac{1}{i+1}&=& \int_0^1 \sum_{i=0}^{n} (-1)^i\binom{n}{i}x^i dx \\ &=& \int_0^1 (1-x)^n dx \\ &=& \left[ \frac{-(1-x)^n}{n+1} \right]^1_0 \\ &=& \color{blue}{ \frac{1}{n+1}} \\ \end{eqnarray*}

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  • $\begingroup$ Your answer is better than mine+1). $\endgroup$ – xpaul Oct 3 '17 at 19:43
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\begin{eqnarray} &&1 - \frac{1}{2}\binom{n}{1}+\frac{1}{3} \binom{n}{2}- \frac{1}{4}\binom{n}{3}+....+(-1)^n \frac{1}{n+1}\binom{n}{n}\\ &=&\sum_{k=0}^n(-1)^{k}\frac{1}{k+1}\binom{n}{k}\\ &=&\sum_{k=0}^n(-1)^{k}\frac{1}{n+1}\binom{n+1}{k+1}\\ &=&\frac{1}{n+1}\sum_{k=1}^{n+1}(-1)^{k}\binom{n+1}{k}\\ &=&-\frac{1}{n+1}\left[\sum_{k=0}^{n+1}(-1)^{k}\binom{n+1}{k}-1\right]\\ &=&\frac1{n+1} \end{eqnarray}

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  • $\begingroup$ $\pm \frac{1}{n+1}$ ... lol ... I was about to change my answer to agree with yours $\ddot \smile$ $\endgroup$ – Donald Splutterwit Oct 3 '17 at 19:42
  • $\begingroup$ Your answer uses the binomial identity we are instructed to use and is therefore probably the solution that is required. $\endgroup$ – Donald Splutterwit Oct 3 '17 at 19:47
  • $\begingroup$ @DonaldSplutterwit, thanks. $\endgroup$ – xpaul Oct 3 '17 at 19:59
  • $\begingroup$ why the last line? The final step? Oh the sum fluctuating combinatoric is 0? what about if n is odd? $\endgroup$ – user4951 Oct 4 '17 at 6:24
  • $\begingroup$ @J.Chang, use $(1-1)^{n+1}=0$. $\endgroup$ – xpaul Oct 4 '17 at 13:52

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