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