Question

Prove the following by using the principle of mathematical induction for all $n\in N.$
$1^2+3^2+5^2+\cdots +(2n-1{)}^2=\frac{n(2n-1)(2n+1)}{3}$

Answer 1

Step $\left(1\right):$ Assume given statement :
Let the given statement be $P\left(n\right)$ i.e.,
$P\left(n\right):1^2+3^2+5^2+\cdots +(2n-1{)}^2=\frac{n(2n-1)(2n+1)}{3}$

Step $\left(2\right):$ Checking statement $P\left(n\right)$ for $n=1$
Put $n=1$ in $P\left(n\right)$, we get
$P\left(1\right):1^2=\frac{1\cdot (2\cdot 1-1)(2\cdot 1+1)}{3}$
$\Rightarrow 1=\frac{1\cdot 1\cdot 3}{3}$
$\Rightarrow 1=1$
Thus $P\left(n\right)$ is true for $n=1$

Step $\left(3\right):$ $P\left(n\right)$ for $n=K$
Put $n=K$in $P\left(n\right)$ and assume this is true for some natural number $K$ i.e.,
$P\left(K\right):1^2+3^2+5^2+\cdots +(2n-1{)}^2=\frac{K(2K-1)(2K+1)}{3}\cdots (i)$

Step $\left(4\right):$ Checking statement $P\left(n\right)$ for $n=K+1$
Now we shall prove that $P(K+1)$ is true whenever $P\left(K\right)$ is true.
Now, we have
$1^2+3^2+5^2+\cdots +(2K-1{)}^2+(2K+1{)}^2$
$=\{1^2+3^2+5^2+\cdots +(2K-1{)}^2\left)\right\}+(2K+1{)}^2$
$=\frac{K(2K-1)(2K+1)}{3}+(2K+1{)}^2$ (using $\left(1\right)$)
$=(2K+1)[\frac{K(2K-1)}{3}+(2K+1\left)\right]$
$=(2K+1)\left[\frac{2K^2-K+6K+3}{3}\right]$
$=(2K+1)\left[\frac{2K^2+5K+3}{3}\right]$
$=(2K+1)\left[\frac{2K^2+3K+2K+3}{3}\right]$
$=(2K+1)\left[\frac{(2K+3)(K+1)}{3}\right]$

We can write it as
$\frac{(K+1)\left\{2\right(K+1)-1\}\left\{2\right(K+1)+1\}}{3}$
Thus, $P(K+1)$ is true whenever $P\left(K\right)$ is true
Final answer :
Therefore, by the principle of mathematical induction, statement $P\left(n\right)$ is true for all $n\in N$.