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How to evaluate the expression $$\displaystyle \sum_{k=1}^n \binom{2k}{2}$$ using a combinatorial argument?

Sorry I have little clue so cannot provide any working for it. Not homework. Related to previous question.

Martin Sleziak
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kuch nahi
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    You can point to the previous question you posted so that others can see the motivation for this question. –  Jun 05 '11 at 05:09

3 Answers3

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A simple combinatorial argument of choosing $2$ balls from $2k$ red balls and $1$ blue ball gives us $$\binom{2k+1}{2} = \binom{2k}{2} + 2k$$ And $$\sum_{k=1}^{2n+1} \binom{k}{2} = \binom{2n+2}{3}$$ Hence $$2 \sum_{k=1}^{n} \binom{2k}{2} + \sum_{k=1}^{n} 2k = \binom{2n+2}{3}$$ and hence $$2 \sum_{k=1}^{n} \binom{2k}{2} = \binom{2n+2}{3} - \sum_{k=1}^{n} 2k$$ $$\sum_{k=1}^{n} \binom{2k}{2} = \frac{\binom{2n+2}{3}}{2} - \binom{n+1}{2}$$

(Some of the machinery for this argument has been borrowed from this question in particular the $\displaystyle \sum_{k=1}^{n} \binom{k}{2} = \binom{n+1}{3}$)

Community
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  • @Sivaram, I am confused here. What is the relationship between $\sum_{k=1}^n \binom{2k+1}{2}$ and $\sum_{k=1}^{2n+1}\binom{k}{2}$ that you have used? – kuch nahi Jun 06 '11 at 17:55
  • I suspect it is $$\sum_{k=1}^n \binom{2k+1}{2} =\frac{1}{2} \sum_{k=1}^{2n+1}\binom{k}{2}$$ from a bijection on $$ \binom{n+1}{r+1} = \binom{n}{r}+\binom{n-1}{r}+\cdots + \binom{r}{r}$$ ?Then an additional factor of 2 would be needed in the second summand of the third equation. Thanks a lot for any clarifications. – kuch nahi Jun 06 '11 at 18:01
  • @yayu: Split the $\sum_{k=1}^{2n+1} \binom{k}{2}$ into odd and even parts and use that $\binom{2k+1}{2} = \binom{2k}{2} + 2k$ for the odd parts –  Jun 06 '11 at 19:12
  • Yes, that is what I had reasoned and tried to say in the second comment. So there should be an additional factor of two (i.e second summation in third equation should be $\sum 4k$ instead of $\sum 2k$) – kuch nahi Jun 06 '11 at 19:30
  • @yayu: Nope only the odd terms contribute $2k$ –  Jun 06 '11 at 19:31
  • I see now. This was a really good answer. (Even better than the last one) – kuch nahi Jun 06 '11 at 19:34
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Well, $\binom{2k}{2} = k(2k-1) = 2k^{2} - k$. So,

$$ \sum_{k=1}^{n} \binom{2k}{2} = 2\sum_{k=1}^{n} k^{2} - \sum_{k=1}^{n} k = n(n+1) \Big( \frac{2n+1}{3} - \frac{1}{2} \Big) $$

Macro
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    yayu is looking for a combinatorial argument. (Though he did not state it in the question explicitly) I have now edited the question to emphasis that he is looking for a combinatorial argument. –  Jun 05 '11 at 05:23
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The magical question:

$$\displaystyle \sum_{k=1}^n \binom{2k}{2}$$

Substitute with the formula:

$$\displaystyle \sum_{k=1}^n \frac{(2k)!}{(2k-2)!2!}$$

Simplify (we can do this because we know for sure that $2k > 1$):

$$\displaystyle \sum_{k=1}^n \frac{(2k)(2k-1)}{2} = 2\sum_{k=1}^n k^2 - \sum_{k=1}^n k$$

And now:

$$2(\frac{n(n+1)(2n+1)}{6}) - \frac{n(n+1)}{2}$$

$$(n+1)\frac{2n(2n+1)-3n}{6}$$

Mateen Ulhaq
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