Question

Show that $\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+....\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}$ using mathematical induction

Answer 1

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

Using principle of mathematical induction

Step 1 $n=1,$ $\text{LHS}=\frac{1}{1-23}=\frac{1}{6}$

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

Step 2

Suppose n = 2 is true,

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

Step 3

Show n = k + 1,

$\left[\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+....+\frac{1}{k\left(k+1\right)\left(k+2\right)}\right]+\frac{1}{\left(k+1\right)\left(k+2\right)\left(k+3\right)}$

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

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

Hence true for n = k.