1.3 + 2.4 + 3.5 + ... + n.

Solution:

Let P(n) be the given statement.

Now,

$P(n)=1.3+2.4+3.5+\ldots+n .(n+2)=\frac{1}{6} n(n+1)(2 n+7)$

Step 1:

 

$P(1)=1.3=3=\frac{1}{6} \times 1(1+1)(2 \times 1+7)$

Hence, $P(1)$ is true.

Step 2 :

 

Let $P(m)$ be true.

Then,

 

1. $3+2 \cdot 4+\ldots+m \cdot(m+2)=\frac{1}{6} m(m+1)(2 m+7)$

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

 

That is,

$1.3+2.4+\ldots+(m+1)(m+3)=\frac{1}{6}(m+1)(m+2)(2 m+9)$

$P(m)$ is equal to $1.3+2.4+\ldots+m(m+2)=\frac{1}{6} m(m+1)(2 m+7)$.

Thus, we have :

 

$1.3+2.4+\ldots+m(m+2)+(m+1)(m+3)=\frac{1}{6} m(m+1)(2 m+7)+(m+1)(m+3)$

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

$\Rightarrow 1.3+2.4+\ldots+(m+1)(m+3)=\frac{1}{6}(m+1)\left[2 m^{2}+7 m+6 m+18\right]$

$=\frac{1}{6}(m+1)\left(2 m^{2}+13 m+18\right)$

 

$=\frac{1}{6}(m+1)(2 m+9)(m+2)$

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

By the $p$ rinciple of $m$ athematical induction, $P(n)$ is true for all $n \in N$.