# Prove the following by using the principle of mathematical induction for all $n in N: 1.2+2.3+3.4+ldots+n .(n+1)=$

Question:

Prove the following by using the principle of mathematical induction for all $n \in N: 1.2+2.3+3.4+\ldots+n .(n+1)=\left[\frac{n(n+1)(n+2)}{3}\right]$

Solution:

Let the given statement be P(n), i.e.,

$\mathrm{P}(n): 1.2+2.3+3.4+\ldots+n \cdot(n+1)=\left[\frac{n(n+1)(n+2)}{3}\right]$

For n = 1, we have

$P(1): 1.2=2=\frac{1(1+1)(1+2)}{3}=\frac{1.2 .3}{3}=2$, which is true.

Let $\mathrm{P}(k)$ be true for some positive integer $k$, i.e.,

$1.2+2.3+3.4+\ldots . .+k \cdot(k+1)=\left[\frac{k(k+1)(k+2)}{3}\right]$ $\ldots$ (i)

We shall now prove that P(k + 1) is true.

Consider

$1.2+2.3+3.4+\ldots+k \cdot(k+1)+(k+1) \cdot(k+2)$

$=[1.2+2.3+3.4+\ldots+k \cdot(k+1)]+(k+1) \cdot(k+2)$

$=\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)$ [Using (i)]

$=(k+1)(k+2)\left(\frac{k}{3}+1\right)$

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

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

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.