Question

prove that 1^2+3^2+5^2+...+(2n+1)^2 = (n+1)(2n+1)(2n+3)/3 whenever n is a nonnegative integer.

Answer 1

We will prove it by induction:

1. When $n = 0: (2 \cdot 0 + 1)^2 = 1 = \frac{(0 + 1)(2 \cdot 0 + 1)(2 \cdot 0 + 3)}{3} \Rightarrow \text{True}$.

2. Let take that if $n = k$ then the given formula is also true. Then we will have:

$$n = k: \sum_{n=0}^{k} (2 \cdot n + 1)^2 = \frac{(k + 1)(2 \cdot k + 1)(2 \cdot k + 3)}{3} \Rightarrow \text{True}.$$

3. Let show that if $n = k + 1$ the formula is true too:

$$\begin{aligned} n = k + 1: & \sum_{n=0}^{k+1} (2 \cdot n + 1)^2 = \sum_{n=0}^{k} (2 \cdot n + 1)^2 + (2 \cdot k + 3)^2 = \\ & = \frac{(k + 1)(2 \cdot k + 1)(2 \cdot k + 3)}{3} + (2 \cdot k + 3)^2 = \frac{(2 \cdot k + 3)(2k^2 + 3k + 1 + 6k + 9)}{3} = \\ & = \frac{(2 \cdot k + 3)(2k^2 + 9k + 10)}{3} = \frac{(2 \cdot k + 3)(2 \cdot k + 5)(k + 2)}{3} \Rightarrow \text{True}. \end{aligned}$$

So, we proved that this formula is true for every nonnegative integer $n$.

Answer: Proved.