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

Answer:

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

Step-by-step explanation:

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

$(2\cdot 1 - 1)^{2} = \frac{2\cdot (2\cdot 1 - 1)\cdot (2\cdot 1 + 1)}{3}$

$1 = \frac{1\cdot 1\cdot 3}{3}$

$1 = 1$

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:

$(1^{2}+2^{2}+3^{2}+...+k^{2}) + [2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}$

$\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)}{3} +[2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}$

$\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)+3\cdot [2\cdot (k+1)-1]^{2}}{3} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}$

$k\cdot (2\cdot k -1)\cdot (2\cdot k +1)+3\cdot (2\cdot k +1)^{2} = (k+1)\cdot (2\cdot k +1)\cdot (2\cdot k +3)$

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

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

$2\cdot k^{2}+5\cdot k +3 = (k+1)\cdot (2\cdot k + 3)$

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

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