Answer:
The statement is true is for any [tex]n\in \mathbb{N}[/tex].
Step-by-step explanation:
First, we check the identity for [tex]n = 1[/tex]:
[tex](2\cdot 1 - 1)^{2} = \frac{2\cdot (2\cdot 1 - 1)\cdot (2\cdot 1 + 1)}{3}[/tex]
[tex]1 = \frac{1\cdot 1\cdot 3}{3}[/tex]
[tex]1 = 1[/tex]
The statement is true for [tex]n = 1[/tex].
Then, we have to check that identity is true for [tex]n = k+1[/tex], under the assumption that [tex]n = k[/tex] is true:
[tex](1^{2}+2^{2}+3^{2}+...+k^{2}) + [2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}[/tex]
[tex]\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)}{3} +[2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}[/tex]
[tex]\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)+3\cdot [2\cdot (k+1)-1]^{2}}{3} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}[/tex]
[tex]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)[/tex]
[tex](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)[/tex]
[tex]k\cdot (2\cdot k - 1)+3\cdot (2\cdot k +1) = (k + 1)\cdot (2\cdot k +3)[/tex]
[tex]2\cdot k^{2}+5\cdot k +3 = (k+1)\cdot (2\cdot k + 3)[/tex]
[tex](k+1)\cdot (2\cdot k + 3) = (k+1)\cdot (2\cdot k + 3)[/tex]
Therefore, the statement is true for any [tex]n\in \mathbb{N}[/tex].
