Question

Prove the following by using the principle of mathematical induction for all $n\in N$
$1\cdot 3+3\cdot 5+5\cdot 7+\cdots +(2n-1)(2n+1)$
$=\frac{n(4n^2+6n-1)}{3}$

Answer 1

Step $\left(1\right):$ Assume given statement
Let the given statement be $P\left(n\right)$, i.e.,
$P\left(n\right):1\cdot 3+3\cdot 5+5\cdot 7+\cdots +(2n-1)(2n+1)$
$=\frac{n(4n^2+6n-1)}{3}$

Step $\left(2\right):$ Checking statement $P\left(n\right)$ for $n=1$
Put $n=1$ in $P\left(n\right)$, we get
$P\left(1\right):1\cdot 3=\frac{1\left(4\right(1{)}^2+6\cdot 1-1)}{3}$
$\Rightarrow 3=\frac{(4+6-1)}{3}$
$\Rightarrow 3=\frac{9}{3}$
$\Rightarrow 3=3$
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.,
$P\left(K\right):1\cdot 3+3\cdot 5+5\cdot 7+\cdots +(2K-1)(2K+1)$
$=\frac{K(4K^2+6K-1)}{3}\cdots \left(1\right)$

Step $\left(4\right):$ Checking statement $P\left(n\right)$ for $n=K+1$
Now we shall prove $P(K+1)$ is true whenever $P\left(K\right)$ is true.
Now, we have
$1\cdot 3+3\cdot 5+5\cdot 7+\cdots +(2K-1)(2K+1)+(2K+1)(2K+3)$
$=\frac{K(4K^2+6K-1)}{3}+(2K+1)(2K+3)$ (using $\left(1\right)$)
$=\frac{K(4K^2+6K-1)}{3}+(4K^2+6K+2K+3)$
$=\frac{K(4K^2+6K-1)+3(4K^2+8K+3)}{3}$
$=\frac{4K^3+6K^2-K+12K^2+24K+9}{3}$
So, we get
$=\frac{4K^3+18K^2+23K+9}{3}$
It can be written as
$=\frac{4K^3+14K^2+9K+4K^2+14K+9}{3}$
$=\frac{K(4K^2+14K+9)+1(4K^2+14K+9)}{3}$
$=\frac{(4K^2+14K+9)(K+1)}{3}$
$=\frac{(K+1)\left\{4\right(K+1{)}^2+6(K+1)-1\}}{3}$
Thus, $P(K+1)$ is true whenever $P\left(K\right)$ is true
Final Answer :
Hence, from the principle of mathematical induction, the statement $P\left(n\right)$ is true for all $n\in N$.