Question

Using mathematical induction prove that:

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

Answer 1

$\begin{array}{c}\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+....+\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}\\ here,\\ p\left(1\right):\frac{1}{\mathrm{1.2.3}}=\frac{1}{6}\\ R.H.s=\frac{1.4}{\mathrm{4.2.3}}=\frac{1}{6}\\ for,n=k\\ p\left(k\right)isalsotrue\\ \frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+....+\frac{1}{k\left(k+1\right)\left(k+2\right)}=\frac{k\left(k+3\right)}{4\left(k+1\right)\left(k+2\right)}\\ for,\left(k+1\right)isalsotrue\\ \frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+.....+\frac{1}{k\left(k+1\right)\left(k+2\right)}+\frac{1}{\left(k+1\right)\left(k+2\right)\left(k+3\right)}=\frac{\left(k+1\right)\left(k+4\right)}{4\left(k+2\right)\left(k+3\right)}\\ L.H.S=\frac{k\left(k+3\right)}{4\left(k+1\right)\left(k+2\right)}+\frac{1}{\left(k+1\right)\left(k+2\right)\left(k+3\right)}\\ =\frac{1}{\left(k+1\right)\left(k+2\right)}\left[\frac{k\left(k+3\right)}{4}+\frac{1}{\left(k+3\right)}\right]\\ =\frac{1}{\left(k+1\right)\left(k+2\right)}\left[\frac{k^3+6k^2+9k+9}{4\left(k+3\right)}\right]\\ =\frac{1}{\left(k+1\right)\left(k+2\right)}.\frac{\left(k+2\right)\left(k+4\right)\left(k+1\right)}{4\left(k+3\right)}\\ =\frac{\left(k+1\right)\left(k+4\right)}{4\left(k+2\right)\left(k+3\right)}\\ =R.H.S\\ so,p\left(k+1\right)istrue\\ p\left(n\right)istrue\forall n\in N.\end{array}$