Using the principle of mathematical induction, prove each of the following

Solution:

To Prove

$1^{2}+3^{2}+5^{2}+7^{2}+\ldots+(2 n-1)^{2}=\frac{n(2 n-1)(2 n+1)}{3}$

Steps to prove by mathematical induction:

Let $P(n)$ be a statement involving the natural number $n$ such that

(i) $P(1)$ is true

(ii) $P(k+1)$ is true, whenever $P(k)$ is true

Then $P(n)$ is true for all $n \in N$

Therefore

Let $P(n): 1^{2}+3^{2}+5^{2}+7^{2}+\ldots+(2 n-1)^{2}=\frac{n(2 n-1)(2 n+1)}{3}$

Step 1:

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

Therefore, P(1) is true

Step 2:

Let P(k) is true Then

$\mathrm{P}(\mathrm{k}): 1^{2}+3^{2}+5^{2}+7^{2}+\ldots+(2 \mathrm{k}-1)^{2}=\frac{k(2 k-1)(2 k+1)}{3}$

Now,

$1^{2}+3^{2}+5^{2}+7^{2}+\ldots+(2(k+1)-1)^{2}=\frac{k(2 k-1)(2 k+1)}{3}+(2 k+2-1)^{2}$

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

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

$=(2 k+1)\left[\frac{2 k^{2}-k+6 k+3}{3}\right]$

$=(2 k+1)\left[\frac{2 k^{2}+5 k+3}{3}\right]$

$={ }^{(2 k+1)\left[\frac{(k+1)(2 k+3)}{3}\right]}$ (Splitting the middle term)

$=\frac{(k+1)(2 k+1)(2 k+3)}{3}$

= P(k + 1)

Hence, P(k + 1) is true whenever P(k) is true

Hence, by the principle of mathematical induction, we have

$1^{2}+3^{2}+5^{2}+7^{2}+\ldots+(2 n-1)^{2}=\frac{n(2 n-1)(2 n+1)}{3}$ for all $n \in N$