Prove a sum formula by induction I am to prove through induction that $$\sum_{k=1}^n (2k-1)^2 = \frac{n(2n-1)(2n+1)}{3}$$
And well, my method seems to be working, but I get stuck when I'm nearly done. First I prove the formula work for $n = 1$.
$$\sum_{k=1}^1 (2k-1)^2 = 1$$
$$f(1)=\frac{(2*1-1)(2*1+1)}{3} = 1$$
I then assume it works for $n = p$ and want to show it works for $n = (p+1)$
$$\frac{n(2n-1)(2n+1)}{3} + (2(n+1)-1)^2 \stackrel{?}{=} \frac{(n+1)(2(n+1)-1)(2(n+1)+1)}{3}$$
$$\frac{n(2n-1)(2n+1)}{3} + (2(n+1)-1)^2 = \frac{(4n^3+12n^2+11n+3)}{3}$$
I didn't show all the steps in the last equation, however, this is where I'm stuck; how can I factor the numerator? I tried using Wolfram | Mathematica, but it just went straight from the current expression into factored form.
Also I suppose I'm wondering if I chose the easiest path to proving the formula through induction.

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



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*$$\sum_{k=1}^1 (2k-1)^2=1^2 = \frac{1(2*1-1)(2*1+1)}{3}=1$$ True

*Assuming it is true for $k=n$:
$$\sum_{k=1}^n (2k-1)^2 = \frac{n(2n-1)(2n+1)}{3}$$ Assumption of Truth

*Now finding for $k=n+1$:
$$\begin{align}\sum_{k=1}^{n+1} (2k-1)^2&=\sum_{k=1}^n (2k-1)^2+(2*(n+1)-1)^2\\&=\frac{n(2n-1)(2n+1)}{3}+(2n+1)^2\\&=\frac{(2n+1)}3(2n^2-n+6n+3)\\&=\frac{(2n+1)}3(2n^2+5n+3)\\&=\frac{(n+1)(2n+1)(2n+3)}{3}\\&=\frac{(n+1)(2*(n+1)-1)(2*(n+1)+1)}{3}\end{align}$$ Establishing the Chain
