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Consider $n\in\mathbb{N}.$ Find the sum of:$$S=\left(\dfrac{C_n^0}{1} \right)^2+\left(\dfrac{C_n^1}{2} \right)^2+\cdots+\left( \dfrac{C_n^n}{n+1}\right)^2$$

I don't know how to solve it, i don't have any ideas, so i am very happy to hear from you. Thank you very much.

I Love Math
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    What is the notation $C_n^{j}$? – Clarinetist Jun 25 '14 at 00:28
  • I think it is $C^1_n$ not $C^j_n$ :) – I Love Math Jun 25 '14 at 00:35
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    I think $C^j_n = {n \choose j}$, no? – Tom Jun 25 '14 at 00:40
  • I really understand, can you make it more cleary, I am very thank you. – I Love Math Jun 25 '14 at 00:42
  • You manually added spacing, writing "+,...,+". But spacing is added automatically when using \dots or \ldots or \cdots, including spacing between the dots. If you write "..." when using LaTeX in the usual way (as opposed to using it on web sites in the form of MathJax and the like), then when you write a,...,b you see $a,\text{...},b$ instead of $a,\ldots,b$ (which is coded as a,\ldots,b) or $a+\text{...}+b$ instead of $a+\cdots+b$. I changed it to "+\cdots+$. ${}\qquad{}$ – Michael Hardy Jun 25 '14 at 01:14
  • Thinking along the lines of integrating $(1+x)^n$, then dividing by $x$ and then integrating again. The value of the integral at $x=1$ should give the required sum. But running into some errors and short of time, so I'll just leave this here as an avenue to explore. – Deepak Jun 25 '14 at 01:24
  • Is the 0 supposed to be below the 'n' in the choose function? –  Jun 25 '14 at 05:35

1 Answers1

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Outline: $$ \frac{{n \choose k}}{k+1} =\frac{n!}{k! (n-k)!(k+1)}=\frac{n!}{(k+1)!(n-k)!} = \frac{1}{n+1}\cdot {n+1 \choose k+1} $$ so $$ \sum_{k=0}^n \Big(\frac{{n \choose k}}{k+1}\Big)^2 = \Big(\frac{1}{n+1}\Big)^2 \sum_{k=0}^n {n+1 \choose k+1}^2 $$

Now, it is known that $$ \sum_{l=0}^{m} {m \choose l}^2 = {2m \choose m} $$ (you can do this inductively, see here). From here... can you finish?

Community
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Tom
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  • I don't know what does the $\left(n\choose k\right)$ mean, can you help me to explanit it, thank you. – I Love Math Jun 25 '14 at 03:46
  • ${n \choose k} = \frac{n!}{k! (n-k)!}$ which is what I believe you mean by $C^k_n$, right? – Tom Jun 25 '14 at 13:49