Prove a sum formula by induction

Question

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.

Answer 1

For $n+1$ step, you don't need to expand them. $$\begin{align}\sum_{k=1}^{n+1}(2k-1)^2&=\sum_{k=1}^{n}(2k-1)^2+\{2(n+1)-1\}^2\\&=\frac{n(2n-1)(2n+1)}{3}+(2n+1)^2\\&=\frac{n(2n-1)\color{red}{(2n+1)}}{\color{red}{3}}+(2n+1)\times \color{red}{(2n+1)}\times \frac{3}{\color{red}{3}}\\&=\color{red}{\frac{2n+1}{3}}\left\{n(2n-1)+3(2n+1)\right\}\\&=\frac{2n+1}{3}(2n^2+5n+3)\\&=\frac{2n+1}{3}(n+1)(2n+3)\\&=\frac{(n+1)\{2(n+1)-1\}\{2(n+1)+1\}}{3}.\end{align}$$

Answer 2

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

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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

Answer 3

How to factor the numerator: $$4n^3 + 12n^2 + 11n+3 = 4n^3 + 8n^2 + 3n + 4n^2 + 8n + 3 = n(4n^2 +8n+3) + (4n^2 +8n + 3) = (n+1)(4n^2 + 8n + 3) = (n+ 1)(2n+1)(2n+3) = (n+1)(2(n+1)-1)(2(n+1)+1)$$