Question

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

Answer 1

Let the given statement be $P\left(n\right)$ i.e.,
$1.2+2.3+3.4+......+n(n+1)=\left[\frac{n(n+1)(n+2)}{3}\right]$
$P\left(n\right)$:
For $n=1$, we have
$1.2=2\frac{1(1+1)(1+2)}{3}=\frac{\mathrm{1.2.3}}{3}=2$, which is true.
Let $P\left(k\right)$ be true for some positive integer $k$ i.e.,
$1.2+2.3+3.4+.....+k.(k+1)=\left[\frac{k(k+1)(k+2)}{3}\right]........\left(i\right)$
We shall now prove that $P(k+1)$ is true.
Consider
$1.2+2.3+3.4+......+k.(k+1)+(k+1)+(k+1).(k+2)$
$=[1.2+2.3+3.4+.....+k.(k+1\left)\right]+(k+1).(k+2)$
$=\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)$ [Using $\left(i\right)$]
$=(k+1)(k+2)(\frac{k}{3}+1)$
$=\frac{(k+1)(k+2)(k+3)}{3}$
$=\frac{(k+1)(k+1+1)(k+1+2)}{3}$
Thus, $P(k+1)$ is true whenever $P\left(k\right)$ is true.
Hence by the principle of mathematical induction, statement $P\left(n\right)$ is true for all natural numbers i.e., $n$