How to use binomial properties when proving by induction?

Question

Can't seem to find this anywhere, so would really appreciate the help. Looking for an explanation rather than just an answer.

For $n ≥ 1$, verify that $1^2$ + $3^2$ + $5^2$ + ... + $(2n-1)^2$ = $\binom{2n+1}{3}$

I've tried looking at other examples but I don't understand how the left-hand side could be transformed to a binomial at $n+1$ level.

Answer 1

If you already suspect it sums to $\binom{2n+1}{3}$, induction should be quite straightforward. To find the formula in the first place it's often useful to try small cases.

Answer 2

$1^2 + 3^2 + ... + (2k - 1)^2= ^{2k - 1}C_3$

$^{2k - 1}C_3 + (2k + 1)^2 = \frac{(2k - 1)(2k - 2)(2k -3)}{3!} + (2k + 1)^2$

Simplify the RHS to $\frac{(2k + 1)(2k)(2k - 1)}{3!} = ^{2k + 1}C_3$

In short, formula true for k and k + 1

Answer 3

by induction hypothesis, \begin{equation*} \begin{split} 1^2+3^2+5^2+\dots+(2k-1)^2&={2n+1 \choose 3}\\ & = \frac{(2k+1)!}{3!(2k-2)!}\\ &=\frac{(2k+1)(2k)(2k-1)}{6} \end{split} \end{equation*}

now for $n=k+1 $, \begin{equation*} \begin{split} 1^2+3^2+5^2+\dots+(2k-1)^2+(2k+1)^2&={2n+1 \choose 3} + (2k+1)^2\\&=\frac{(2k+1)(2k)(2k-1)}{6} + (2k+1)^2 \\&= \frac{(2k+1)(2k+2)(2k+3)}{6} \\ &=\frac{(2k+3)!}{3!(2k)!} \\&={2k+3 \choose 3}\\&= {2(k+1)+1 \choose 3} \end{split} \end{equation*} Hence Proved.