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It is showable directly by induction that the following are true:

$$\sum k = \frac{1}{2}n(n+1)$$

$$\sum k^2 = \frac{1}{6}n(n+1)(2n+1)$$

$$\sum k^3 = \frac{1}{4}n^2(n+1)^2$$

etc. Now, by doing some minor rearranging, we can make a new pattern appear. Combining the first two equations, we can observe that.

$$\sum k(k+1) = \sum k^2 + k = \sum k^2 + \sum k = \frac{1}{3}(n)(n+1)(n+2)$$

after some algebraic rearrangements.

Similarly, if we're patient, we can see that $$\sum k(k+1)(k+2) = \frac{1}{4}n(n+1)(n+2)(n+3)$$

It's not hard to imagine that this pattern continues, and in fact a quick google search shows that it does continue. The general formula is $$ \sum \limits_{k=1}^{n} k(k+1)(k+2)...(k+p-1) = \frac{1}{p+1}n(n+1)...(n+p)$$

However, the only way to prove it I can think of is difficult. It involves first finding a formula for an arbitrary sum of powers, and then doing some very ugly rearranging and summing known formulas like in these first two cases. Is there some more obvious or more direct way of proving it?

Edit: The bot is telling me there is a similar question. This is true. In the other question, the answers refer to either a case by case analysis, which I am aware of how to do. In one answer, someone uses a calculus based proof. I'd like to see an algebraic proof if that's possible.

A. Thomas Yerger
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  • Title edited. Sorry about that. – A. Thomas Yerger Dec 24 '15 at 18:28
  • Welp. I'm not doing very good today @NormalHuman. So then what to do now that there is an answer? Put a bounty on the existing thread and have the other answerer repost there? Take to meta? – A. Thomas Yerger Dec 24 '15 at 18:49
  • I think http://math.stackexchange.com/q/219693 is the best target for this. I voted to close others as its duplicate. –  Dec 24 '15 at 20:27

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We have $$k(k+1)\cdots(k+p-1) = \dfrac{(k+p-1)!}{(k-1)!} = p! \dbinom{k+p-1}{p}$$ Now we have $$\sum_{k=1}^n \dbinom{k+p-1}{p} = \dbinom{n+p}{p+1}$$ which follows from the fact that $\dbinom{n+1}r = \dbinom{n}r + \dbinom{n}{r-1}$. This proves what you want.

Adhvaitha
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  • This is really simple! I was thinking waaay too hard about this problem. – A. Thomas Yerger Dec 24 '15 at 18:45
  • @NormalHuman the sum of the two binomial coefficients on the right? That can be done just be expanding both into their factorial expressions and combining the fractions. – A. Thomas Yerger Dec 24 '15 at 18:54
  • @AlfredYerger Yes, but how does $\sum_{k=1}^n \binom{k+p-1}{p} = \binom{n+p}{p+1}$ follow from that? –  Dec 24 '15 at 18:56
  • It's proven by induction. clearer I think when you think of the sum as running over $\dbinom{p+(k-1)}{p}$. – A. Thomas Yerger Dec 24 '15 at 19:05
  • @AlfredYerger If Leg doesn't want to explain this step, perhaps you will, with an edit? So that we have a clear, complete, canonical answer to this question. –  Dec 24 '15 at 19:17
  • I don't think a proof will fit into a comment. Should I edit it into the post? I think also I recall seeing a proof of that claim on proofwiki - I could probably find it and drop a link. – A. Thomas Yerger Dec 24 '15 at 19:19