Question

Prove the following by using the principle of mathematical induction for all $n\in N:\mathrm{1.2.3}+\mathrm{2.3.4}+......+n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4}$

Answer 1

Let the given statement be $P\left(n\right)$, i.e.,
$P\left(n\right):\mathrm{1.2.3}+\mathrm{2.3.4}+.....+n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4}$
For $n=1$, we have
$P\left(1\right):\mathrm{1.2.3}=6=\frac{1(1+1)(1+2)(1+3)}{4}=\frac{\mathrm{1.2.3.4}}{4}=6$, which is true.
Let $P\left(k\right)$ be true for some positive integer $k$ i.e.,
$\mathrm{1.2.3}+\mathrm{2.3.4}+.....k(k+1)(k+2)=\frac{k(k+1)(k+2)(k+3)}{4}..........\left(i\right)$
We shall now prove that $P(k+1)$ is true.
Consider
$\mathrm{1.2.3}+\mathrm{2.3.4}+......+k(k+1)(k+2)+(k+1)(k+2)(k+3)=$

$1,2,3+2,3,4+.....k(k+1)(k+2)+(k+1)(k+2)(k+3)$
$=\frac{k(k+1)(k+2)(k+3)}{4}+(k+1)(k+2)(k+3)$ [Using $\left(i\right)$]
$=(k+1)(k+2)(k+3)(\frac{k}{4}+1)$
$=\frac{(k+1)(k+2)(k+3)(k+4)}{4}$
$=\frac{(k+1)(k+1+1)(k+1+2)(k+1+3)}{4}$
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., $n$