Question

Prove by using the principle of mathematical induction that $\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{n(n+1)}=\frac{n}{n+1}$

Answer 1

Let P(n) be the statement given by

P(n) : $\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{n(n+1)}=\frac{n}{n+1}$

For n = 1, we get

LHS = $\frac{1}{1.2}=\frac{1}{2}\text{and}\text{RHS}=\frac{1}{1+1}=\frac{1}{2}$

So, P(1) is true,

For n = k, assume P(k) is true.

Then, $\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{k(k+1)}=\frac{k}{k+1}$

We have to show that, P(k+1) is true whenever P(k) is true.

So, add $\frac{1}{(k+1)(k+2)}$ on both sides of Eq. (i), we get

$\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}$

$\frac{k(k+2)+1}{(k+1)(k+2)}=\frac{k^2+2k+1}{(k+1)(k+2)}$

$\frac{(k+1{)}^2}{(k+1)(k+2)}=\frac{k+1}{k+2}$

So, P(k+1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction,

P(n) is true for all n$ϵ$ N.