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