Proof identities (i)$\sum_{k=0}^{n}\frac{1}{k+1}\binom{n}{k}=\frac{2^{n+1}-1}{n+1}$ and (ii) $\sum_{k=0}^{m}\binom{n+k}{k}=\binom{n+m+1}{n+1}$

Asked Nov 01 '23 at 03:34

Active Nov 01 '23 at 04:44

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I'm really having trouble with this assignment. You have to prove the following identities for $m, n ∈ \mathbb{N}$ including $0$.

(i)$\sum_{k=0}^{n}\frac{1}{k+1}\binom{n}{k}=\frac{2^{n+1}-1}{n+1}$

(ii)$\sum_{k=0}^{m}\binom{n+k}{k}=\binom{n+m+1}{n+1}$

I can only think of induction. I would set $n=0$ at the beginning of the induction for (i) and then choose an arbitrary but fixed $n ∈ \mathbb{N}$ including 0 as a condition of induction, where (i) holds. However, I do not get further with the induction step. I replace $n \to n+1$ but then what? I also thought about multiplying by $n+1$ on both sides, but I can't reach the conclusion. With (ii) I do not get further either. Can someone explain to me how to proceed here?

edited Nov 01 '23 at 04:44

cpiegore

1,490

asked Nov 01 '23 at 03:34

ArtanR

1

For (i): https://math.stackexchange.com/questions/1706128/binomial-theorem-sum – Minus One-Twelfth Nov 01 '23 at 03:46
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For (ii): https://math.stackexchange.com/questions/357063/proving-that-sum-limits-k-0n-mk-choosem-mn1-choose-m1 (Note: this identity is often called the "hockey stick identity", so you can search for that to find more information about it) – Minus One-Twelfth Nov 01 '23 at 03:53

Proof identities (i)$\sum_{k=0}^{n}\frac{1}{k+1}\binom{n}{k}=\frac{2^{n+1}-1}{n+1}$ and (ii) $\sum_{k=0}^{m}\binom{n+k}{k}=\binom{n+m+1}{n+1}$

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