Question

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

Answer 1

Let us assume that,

$p\left(x\right)=\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+\cdots +\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}$

for $n=1$

L.H.S $=\frac{1}{\mathrm{1.2.3}}=\frac{1}{6}$

R.H.S $=\frac{1.\left(1+3\right)}{4\left(1+1\right)\left(1+2\right)}=\frac{1}{6}$

$P\left(1\right)$ is true.

Let, $P\left(k\right)$ is true

i.e,

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

To show :- $P(k+1)$ is true

Now,

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

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

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

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

$=\frac{1}{\left(k+1\right)\left(k+2\right)}\frac{k\left(k^2+6k+9\right)+4}{4\left(k+3\right)}$

$=\frac{1}{\left(k+1\right)\left(k+2\right)}\frac{k^3+6k^2+9k+4}{4\left(k+3\right)}$

$=\frac{1}{\left(k+1\right)\left(k+2\right)}\frac{k^3+k^2+5k^2+4k4}{4\left(k+3\right)}$

$=\frac{k^2\left(k+1\right)+5k\left(k+1\right)+4\left(k+1\right)}{4\left(k+1\right)\left(k+2\right)\left(k+3\right)}$

$=\frac{\left(k+1\right)\left(k^2+5k+4\right)}{4\left(k+1\right)\left(k+2\right)\left(k+3\right)}$

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

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

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