Question

Using the principle of mathematical induction, prove that
$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{n(n+1)}=\frac{n}{n+1}$, for all $n\in N$

Answer 1

Let
$P\left(n\right)=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{n(n+1)}=\frac{n}{n+1}$
For $n=1,LHS=\frac{1}{1.2}=\frac{1}{2};RHS=\frac{1}{1+1}=\frac{1}{2}$
$\therefore LHS=RHS$
$\therefore P\left(1\right)$ is true
Let us assume $P\left(k\right)$ is true for some $k\in N$
i..e, $\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{k(k+1)}=\frac{k}{k+1}$
Adding $(k+1{)}^{th}$ term $=\frac{1}{(k+1)(k+2)}$ on both sides,
we get,
$\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}$
$=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}$
$=\frac{k(k+2)+1}{(k+1)(k+2)}$
$=\frac{k^2+2k+1}{(k+1)(k+2)}$
$=\frac{k+1{)}^2}{k+1\left)\right(k+2)}=\frac{k+1}{k+2}$
$=\frac{k+1}{k+1)+1}$
which is $P(k+1)$
Thus, $P\left(k\right)\Rightarrow P(k+1)$
Hence, by mathematical induction $P\left(n\right)$ is true for all $n\in N$