Question

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

Answer 1

$\begin{array}{c}P\left(n\right):\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}\\ forn=1,wehave\\ P\left(1\right):\frac{1}{\mathrm{1.2.3}}=\frac{1\left(1+3\right)}{4\left(1+1\right)\left(1+2\right)}=\frac{1.4}{\mathrm{4.2.3}}=\frac{1}{\mathrm{1.2.3}},whichistrue\\ Let,P\left(k\right)betrueforsomepositiveintegerk,i.e.,\\ P\left(k\right):\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+...+\frac{1}{k\left(k+1\right)\left(k+2\right)}=\frac{k\left(k+3\right)}{4\left(k+1\right)\left(k+2\right)}.........\left(1\right)\\ weshallnowprovethatP(k+1)istrue\\ \\ consider\\ P(k+1):\left[\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+...+\frac{1}{k\left(k+1\right)\left(k+2\right)}\right]+\frac{1}{\left(k+1\right)\left(k+2\right)\left(k+3\right)}\\ =\frac{k\left(k+3\right)}{4\left(k+1\right)\left(k+2\right)}+\frac{1}{\left(k+1\right)\left(k+2\right)\left(k+3\right)}.................u\sin g\left(1\right)\\ =\frac{1}{\left(k+1\right)\left(k+2\right)}\left\{\frac{k\left(k+3\right)}{4}+\frac{1}{\left(k+3\right)}\right\}\end{array}$
$\begin{array}{c}=\frac{1}{\left(k+1\right)\left(k+2\right)}\left\{\frac{k{\left(k+3\right)}^2+4}{4\left(k+3\right)}\right\}\\ =\frac{1}{\left(k+1\right)\left(k+2\right)}\left\{\frac{k\left(k^2+9+6k\right)+4}{4\left(k+3\right)}\right\}\\ =\frac{1}{\left(k+1\right)\left(k+2\right)}\left\{\frac{k^3+9k+6k^2+4}{4\left(k+3\right)}\right\}\\ =\frac{1}{\left(k+1\right)\left(k+2\right)}\left\{\frac{k^3+2k^2+k+4k^2+8k+4}{4\left(k+3\right)}\right\}\\ =\frac{1}{\left(k+1\right)\left(k+2\right)}\left\{\frac{k(k^2+2k+1)+4(k^2+2k+1)}{4\left(k+3\right)}\right\}\\ =\frac{1}{\left(k+1\right)\left(k+2\right)}\left\{\frac{k{(k+1)}^2+4{(k+1)}^2}{4\left(k+3\right)}\right\}\\ =\frac{{(k+1)}^2(k+4)}{4\left(k+3\right)\left(k+1\right)\left(k+2\right)}\\ =\frac{(k+1)\left\{(k+1)+3\right\}}{4\left\{(k+1)+1\right\}\left\{(k+1)+2\right\}}\\ Thus,P(k+1),istruewheneverP\left(k\right)istrue\end{array}$