Question

Use mathematical induction to prove the statement is true for all positive integers n. 1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = (n(2n-1)(2n+1))/3)

Answer 1

The statement is true is for any
`n\in \mathbb{N}`.

Explanation:

First, we check the identity for
`n = 1`:

`(2\cdot 1 - 1)^(2) = (2\cdot (2\cdot 1 - 1)\cdot (2\cdot 1 + 1))/(3)`

`1 = (1\cdot 1\cdot 3)/(3)`

`1 = 1`

The statement is true for
`n = 1`.

Then, we have to check that identity is true for
`n = k+1`, under the assumption that
`n = k` is true:

`(1^(2)+2^(2)+3^(2)+...+k^(2)) + [2\cdot (k+1)-1]^(2) = ((k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1])/(3)`

`(k\cdot (2\cdot k -1)\cdot (2\cdot k +1))/(3) +[2\cdot (k+1)-1]^(2) = ((k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1])/(3)`

`(k\cdot (2\cdot k -1)\cdot (2\cdot k +1)+3\cdot [2\cdot (k+1)-1]^(2))/(3) = ((k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1])/(3)`

`k\cdot (2\cdot k -1)\cdot (2\cdot k +1)+3\cdot (2\cdot k +1)^(2) = (k+1)\cdot (2\cdot k +1)\cdot (2\cdot k +3)`

`(2\cdot k +1)\cdot [k\cdot (2\cdot k -1)+3\cdot (2\cdot k +1)] = (k+1) \cdot (2\cdot k +1)\cdot (2\cdot k +3)`

`k\cdot (2\cdot k - 1)+3\cdot (2\cdot k +1) = (k + 1)\cdot (2\cdot k +3)`

`2\cdot k^(2)+5\cdot k +3 = (k+1)\cdot (2\cdot k + 3)`

`(k+1)\cdot (2\cdot k + 3) = (k+1)\cdot (2\cdot k + 3)`

Therefore, the statement is true for any
`n\in \mathbb{N}`.