Question

Using the principle of mathematical induction, prove that

$\mathrm{1.2.3}+\mathrm{2.3.4}+\mathrm{3.4.5}+...+n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4},$ for all $n\in N.$

Answer 1

Let P(n) be the statement given by

$P\left(n\right):\mathrm{1.2.3}+\mathrm{2.3.4}+...+n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4}$

Put $n=1$

LHS: $P\left(1\right)=\mathrm{1.2.3}=6$
RHS : $P\left(1\right)=\frac{1(1+1)(1+2)(1+3)}{4}=\frac{1\times 2\times 3\times 4}{4}=6$

So, P(1) is true. Now, assume that, P(k) is true, for some natural number k, such that

$\mathrm{1.2.3}+\mathrm{2.3.4}+...+k(k+1)(k+2)=\frac{k(k+1)(k+2)(k+3)}{4}...\left(i\right)$

We will now to show that $P(k+1)$ is true, whenever P(k) is true.

On adding $(k+1),(k+2)(k+3)$ to both sides of eq. (i), we get

$\mathrm{1.2.3}+\mathrm{2.3.4}+...+k(k+1)(k+2)+(k+1)(k+2)(k+3)$

= $\frac{k(k+1)(k+2)(k+3)}{4}+(k+1)(k+2)(k+3)$ [using Eq. (i)]

= $(k+1)(k+2)(k+3)(\frac{k}{4}+1)=\frac{(k+1)(k+2)(k+3)(k+4)}{4}=\frac{(k+1)\left(\right(k+1)+1)\left(\right(k+1)+2)\left(\right(k+1)+3)}{4}$

So, $P(k+1)$ is true.

Hence, by principle of mathematical induction, P(n) is true, $\forall n\in N$