Using mathematical induction prove that ${{1}^{2}}+{{3}^{2}}+{{5}^{2}}...........{{(2n-1)}^{2}}=\dfrac{n\left( 2n-1 \right)\left( 2n+1 \right)}{3}\forall n\in N$

Hint: First, start by checking the domain of the inequality followed by using the method of mathematical induction to find the answer.
We are asked to prove the relation given in the figure using mathematical induction and according to the rule of mathematical induction we first check the relation for the base value of the variable, followed by considering the relation to be true for a value k and using this we try to reach to a result which makes the relation true for the value k+1.

Complete step-by-step answer:
So, according to the rule of mathematical induction:
We will check if the relation to be true for the base value, i.e., n=1 and consider the relationship to be true for n=k and then if using this relation, we could prove that the relation holds for n=k+1 then we will say that we have proved the required thing.
Let’s start by taking n=1.
${{1}^{2}}=\dfrac{1\times (2-1)(2+1)}{3}$
$\Rightarrow 1=1$
So, the relationship is satisfied. Now let the relation be true for n=k.
${{1}^{2}}+{{3}^{2}}+{{5}^{2}}...........{{(2k-1)}^{2}}=\dfrac{k\left( 2k-1 \right)\left( 2k+1 \right)}{3}$
Now we will add ${{\left( 2k+1 \right)}^{2}}$ to both sides of the equation. On doing so, we get
${{1}^{2}}+{{3}^{2}}+{{5}^{2}}...........{{(2k-1)}^{2}}+{{\left( 2k+1 \right)}^{2}}=\dfrac{k\left( 2k-1 \right)\left( 2k+1 \right)}{3}+{{\left( 2k+1 \right)}^{2}}$
$\Rightarrow {{1}^{2}}+{{3}^{2}}+{{5}^{2}}...........{{(2k-1)}^{2}}+{{\left( 2k+1 \right)}^{2}}=\left( 2k+1 \right)\left( \dfrac{k\left( 2k-1 \right)}{3}+2k+1 \right)$
Now, if we take 3 as the LCM of the right hand side of the equation, we get
${{1}^{2}}+{{3}^{2}}+{{5}^{2}}...........{{(2k-1)}^{2}}+{{\left( 2k+1 \right)}^{2}}=\left( 2k+1 \right)\left( \dfrac{k\left( 2k-1 \right)+6k+3}{3} \right)$
$\Rightarrow {{1}^{2}}+{{3}^{2}}+{{5}^{2}}...........{{(2k-1)}^{2}}+{{\left( 2k+1 \right)}^{2}}=\left( 2k+1 \right)\left( \dfrac{2{{k}^{2}}-k+6k+3}{3} \right)$
$\Rightarrow {{1}^{2}}+{{3}^{2}}+{{5}^{2}}...........{{(2k-1)}^{2}}+{{\left( 2k+1 \right)}^{2}}=\left( 2k+1 \right)\left( \dfrac{2{{k}^{2}}+5k+3}{3} \right)$
Now we know we can break 5k as 2k+3k. So, on doing so, we get
${{1}^{2}}+{{3}^{2}}+{{5}^{2}}...........{{(2k-1)}^{2}}+{{\left( 2k+1 \right)}^{2}}=\left( 2k+1 \right)\left( \dfrac{2{{k}^{2}}+2k+3k+3}{3} \right)$
$\Rightarrow {{1}^{2}}+{{3}^{2}}+{{5}^{2}}...........{{(2k-1)}^{2}}+{{\left( 2k+1 \right)}^{2}}=\left( 2k+1 \right)\left( \dfrac{\left( 2k+3 \right)\left( k+1 \right)}{3} \right)$
$\Rightarrow {{1}^{2}}+{{3}^{2}}+{{5}^{2}}...........{{(2k-1)}^{2}}+{{\left( 2k+1 \right)}^{2}}=\dfrac{\left( k+1 \right)\left( 2k+1 \right)\left( 2k+3 \right)}{3}$
The final equation we got is the relation given in the question with the value of n equal to k+1.
So, we have shown that the relation holds true for n=k+1 as well. Hence, by the rule of mathematical induction, we can say that ${{1}^{2}}+{{3}^{2}}+{{5}^{2}}...........{{(2n-1)}^{2}}=\dfrac{n\left( 2n-1 \right)\left( 2n+1 \right)}{3}\forall n\in N$ .

Note: Don’t get afraid looking at the long terms given in the equations, as in questions related to mathematical induction, the expressions looks to be really long ,but in actuality most of the terms are not to be touched and are just to be kept as it is while solving the questions, as we saw in the above question.