Question

Prove the following by using the principle of mathematical induction for all $n\in N:\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+.....+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$

Answer 1

Let the given statement be $P\left(n\right)$ i.e.,
$P\left(n\right):\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+.....+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$
For $n=1$, we have
$P\left(1\right):\frac{1}{\mathrm{1.2.3}}=\frac{1.(1+3)}{4(1+1)(1+2)}=\frac{1.4}{\mathrm{4.2.3}}=\frac{1}{\mathrm{1.2.3}}$, which is true.
Let $P\left(k\right)$ be true for some $k\in N$ i.e.,
$\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+.....+\frac{1}{k(k+1)(k+2)}=\frac{k(k+3)}{4(k+1)(k+2)}........\left(i\right)$
We shall now prove that $P(k+1)$ is true.
$\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+.....+\frac{1}{k(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}=\frac{(k+1)(k+4)}{4(k+2)(k+3)}$
Consider
$[\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+.....+\frac{1}{k(k+1)(k+2)}]+\frac{1}{(k+1)(k+2)(k+3)}$
$=\frac{1}{(k+1)(k+2)}\{\frac{k(k+3)}{4}+\frac{1}{k+3}\}$ [Using $\left(i\right)$]
$=\frac{1}{(k+1)(k+2)}\left\{\frac{k{(k+3)}^2+4}{4(k+3)}\right\}$
$=\frac{1}{(k+1)(k+2)}\left\{\frac{k(k^2+6k+9)+4}{4(k+3)}\right\}$
$=\frac{1}{(k+1)(k+2)}\left\{\frac{k^3+6k^2+9k+4}{4(k+3)}\right\}$
$=\frac{1}{(k+1)(k+2)}\left\{\frac{k^3+k^2+5k^2+5k+4k+4}{4(k+3)}\right\}$
$=\frac{1}{(k+1)(k+2)}\left\{\frac{(k+1)(k^2+5k+4)}{4(k+3)}\right\}$
$=\frac{1}{(k+1)(k+2)}\left\{\frac{(k+1{)}^2(k+4)}{4(k+3)}\right\}$
$=\frac{(k+4)(k+1)}{4(k+1)(k+2)(k+3)}$
$=\frac{(k+1)\left\{\right(k+1)+3\}}{4\left\{\right(k+1)+1\}\left\{\right(k+1)+2\}}$
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., $n$