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Induction for $1^2+3^2+5^2+...+(2n-1)^2= \binom{2n+1}{3}$.

I am stuck... This is what I have so far... Base case: $n=1, 1=\binom{2n+1}{3}\rightarrow 1=1$.

Inductive step: Assume $n=k$ is true and show $n=k+1$ holds.

$$1^2+3^2+5^2+...+(2k-1)^2+(2(k+1)-1)^2= \binom{2(k+1)+1}{3}$$ $$\binom{2k+1}{3}+(2(k+1)-1)^2=\binom{2k+2}{3}$$ $$\frac{(2k+1)!}{(2k+1-3)!3!}+(2(k+1)-1)^2=\binom{2(k+1)+1}{3}$$ $$\frac{(2k+1)!}{(2k-2)!3!}+(2(k+1)-1)^2=\binom{2(k+1)+1}{3}$$

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MathIsHard
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  • Well... can your prove $4k^2 = \frac{(2k+2)!}{(2k-1)!3!} - \frac{(2k+1)!}{(2k-2)!3!}$? Try putting them over a common denominator? $\frac{(2k+1)! - (2k+1)!(2k-1)}{(2k-1)!3!}$? – fleablood Jan 27 '17 at 02:16
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    You need to be careful. Your line $$1^2+3^2+...+(2n-1)^2+(2n+1-1)^2$$

    should read

    $$1^2+3^2+...+(2n-1)^2+(2(n+1)-1)^2$$

    which is equal to

    $$1^2+3^2+...+(2n-1)^2+(2n+2-1)^2$$

     – JSP Jan 27 '17 at 02:18

1 Answers1

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$$[1^2 + \cdots + (2k-1)^2]+ (2(k+1)-1)^2=\binom{2k+1}{3} + (2k+1)^2=\frac{(2k+1)(2k)(2k-1) + 24k^2 + 24k + 6}{6}=\frac{8k^3 + 24k^2+22k+6}{6}=\frac{(2k+3)(2k+2)(2k+1)}{6}=\binom{2(k+1)+1}{3}.$$

Maffred
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  • Thank you for the help. I am confused how to simplify the n choose k into something helpful. I looked up that it turns into an expression with factorials but is that helpful for this problem? – MathIsHard Jan 27 '17 at 02:33
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    @Math4Life Yes, you can simplify and get on the upper side a multiplication of decreasing numbers starting from the number on top, which has the number of bottom terms. On the bottom side you have $k!$. For example $\binom{7}{3}=\frac{765}{3!}$. – Maffred Jan 27 '17 at 02:42
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    Thank you so much. I appreciate the help. I think I see it now. Writing it up shortly on paper for myself :) – MathIsHard Jan 27 '17 at 03:45
  • I imagine there should be an intuitive combinatorial argument, but this is all I could find: http://oeis.org/A000447 – Math1000 Jan 27 '17 at 03:49