Question

$\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+\cdots +\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}.$

Answer 1

Let P (n) $=\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+\cdots +\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}.$
$\text{For}n=1P\left(1\right)=\frac{1}{1(1+2)(1+2)}=\frac{1(1+3)}{4(1+1)(1+2)}\Rightarrow \frac{1}{1\times 2\times 3}=\frac{4}{4\times 2\times 3}\Rightarrow frac16=\frac{1}{6}\therefore$ P (1) is true
Let P(n) be true for n = K
$\therefore P\left(k\right)=\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}\cdots +\frac{1}{k(k+1)(k+2)}+\frac{k(k+3)}{4(k+1)(k+2)}$
For n = k + 1
$\text{R.H.S.}=\frac{(k+1)(k+4)}{4(k+2)(k+3)}\text{L.H.S.}=\frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}=\frac{1}{(k+1)(k+2)}[\frac{k^2+3k}{4}+\frac{1}{k+3}]=\frac{1}{(k+1)(k+2)}\left[\frac{k^3+6k^2+9k+4}{4(k+3)}\right]=\frac{1}{(k+1)(k+2)}\left[frac(k+1{)}^2(k+4)4(k+3)\right]=\frac{(k+1)(k+4)}{4(k+2)(k+3)}\therefore$ P (k + 1) is true
thus P (k) is true $\Rightarrow P(k+1)$ is true
Hence by principle of mathematical induction P (n) is true for all $nϵN.$