1.3 + 3.5 + 5.7 + … + (2n − 1) (2n + 1)
Question:

$1.3+3.5+5.7+\ldots+(2 n-1)(2 n+1)=\frac{n\left(4 n^{2}+6 n-1\right)}{3}$

Solution:

Let P(n) be the given statement.

Now,

$P(n)=1.3+3.5+5.7+\ldots+(2 n-1)(2 n+1)=\frac{n\left(4 n^{2}+6 n-1\right)}{3}$Step 1:

Step 1:

$P(1)=1.3=3=\frac{1\left(4 \times(1)^{2}+6 \times 1-1\right)}{3}$

Hence, $P(1)$ is true.

Step 2:

Let $P(m)$ be true.

Then;

$1.3+3.5+\ldots+(2 m-1)(2 m+1)=\frac{m\left(4 m^{2}+6 m-1\right)}{3}$

To prove : $P(m+1)$ is true.

That is,

$1.3+3.5+\ldots+(2 m+1)(2 m+3)=\frac{(m+1)\left[4(m+1)^{2}+6(m+1)-1\right]}{3}$

Now, $P(m)$ is equal to:

$1.3+3.5+\ldots+(2 m-1)(2 m+1)=\frac{m\left(4 m^{2}+6 m-1\right)}{3}$

$\Rightarrow 1.3+3.5+\ldots+(2 m-1)(2 m+1)+(2 m+1)(2 m+3)=\frac{m\left(4 m^{2}+6 m-1\right)}{3}$

$+(2 m+1)(2 m+3) \quad$ [Adding $(2 m+1)(2 m+3)$ to both sides]

$\Rightarrow P(m+1)=\frac{m\left(4 m^{2}+6 m-1\right)+3\left(4 m^{2}+8 m+3\right)}{3}$

$\Rightarrow P(m+1)=\frac{4 m^{3}+6 m^{2}-m+12 m^{2}+24 m+9}{3}=\frac{4 m^{3}+18 m^{2}+23 m+9}{3}$

$\Rightarrow P(m+1)=\frac{4 m\left(m^{2}+2 m+1\right)+10 m^{2}+19 m+9}{3}$

$=\frac{4 m(m+1)^{2}+(10 m+9)(m+1)}{3}$

$=\frac{(m+1)[4 m(m+1)+10 m+9]}{3}$

$=\frac{(m+1)}{3}\left(4 m^{2}+8 m+4+6 m+5\right)$

$=\frac{(m+1)\left[4(m+1)^{2}+6(m+1)-1\right]}{3}$

Thus, $P(m+1)$ is true.

By the principle of $m$ athematical $i$ nduction, $P(n)$ is true for all $n \in N$.