1.2 + 2.3 + 3.4 + … + n (n + 1)
Question:

$1.2+2.3+3.4+\ldots+n(n+1)=\frac{n(n+1)(n+2)}{3}$

Solution:

Let P(n) be the given statement.

Now,

$P(n)=1.2+2.3+3.4+\ldots+n(n+1)=\frac{n(n+1)(n+2)}{3}$

Step 1:

$P(1)=1.2=2=\frac{1(1+1)(1+2)}{3}$

Hence, $P(1)$ is true.

Step 2:

Let $P(m)$ be true.

Then,

$1.2+2.3+\ldots+m(m+1)=\frac{m(m+1)(m+2)}{3}$

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

That is,

$1.2+2.3+\ldots+(m+1)(m+2)=\frac{(m+1)(m+2)(m+3)}{3}$

Now, $P(m)$ is

$1.2+2.3+\ldots+m(m+1)=\frac{m(m+1)(m+2)}{3}$

$\Rightarrow 1.2+2.3+\ldots+m(m+1)+(m+1)(m+2)=\frac{m(m+1)(m+2)}{3}+(m+1)(m+2)$

 

$\Rightarrow P(m+1)=\frac{(m+1)(m+2)(m+3)}{3}$

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

 

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

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