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How can $\sum_{k=i}^{n} \binom{k}{i}$ be written as

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

I cannot figure this one out, I have tried first proving by induction that the first formula is equal $\sum_{k=i}^{n} \binom{k}{i} = \binom {n+1}{i+1}$. But that did not seem to work for me, because I am confused by the fact that the summation index is equal to the lower limit, $k=i$. How could one approach this problem? And what does this equality mean?

Does the equality $$\sum_{k=i}^{n} \binom{k}{i} = \binom {n+1}{i+1}$$ hold because ${n+1\choose i+1}={k\choose i}+{k+1\choose i+1} ? $

MathFail
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Aristarchus_
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  • @RobertZ; @MathFail gave an excellent answer to my initial question. But the induction proof itself would be useful if you happen to know it. $$ \frac {k(k-1)...(k-i+1)}{i!} = \frac {(k+1)(k)(k-1)...(k-i+1)}{(i+1) i!}$$ I came this far, is this satisfactory or am I missing something? – Aristarchus_ Jul 25 '22 at 16:06
  • $\frac {k(k-1)...(k-i+1)}{i!} = \frac {(k+1)(k)(k-1)...(k-i+1)}{(i+1) i!}$ implies $1= \frac {(k+1)}{(i+1)}$ which does not hold generally. – Robert Z Jul 25 '22 at 16:37
  • @RobertZ; What would the inductive proof look like? – Aristarchus_ Jul 25 '22 at 17:03
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    See https://math.stackexchange.com/questions/2529714/proving-that-binomrr-binomr-1r-cdots-binomnr-binomn?noredirect=1 – Robert Z Jul 25 '22 at 17:09

1 Answers1

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As you said, if you could prove:

$$\sum_{k=i}^{n} \binom{k}{i} = \binom {n+1}{i+1}$$

Then we have:

$$\begin{align} \sum_{k=i}^{n} \frac{k!}{i!(k-i)!} &= \frac{(n+1)!}{(i+1)!(n-i)!}\\ \\ \sum_{k=i}^{n} \frac{k(k-1)...(k-i+1)}{i!} &= \frac{(n+1)n...(n-i+1)}{(i+1)!}\\ \\ \frac{1}{i!}\sum_{k=i}^{n} k(k-1)...(k-i+1) &= \frac{(n+1)n...(n-i+1)}{i!\cdot(i+1)}\\ \\ \sum_{k=i}^{n} k(k-1)...(k-i+1) &= \frac{(n+1)n...(n-i+1)}{i+1} \end{align}$$

MathFail
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