Question

With the help of mathematical induction find for all $n\ge 1$ the sum of the series $\frac{1}{1.2}+\frac{1}{2.3}...\frac{1}{n.n+1}$ is equal to

• A. $n+1$
• B. $\frac{n}{n+1}$
• C. $\frac{2n}{n+1}$
• D. $\frac{n}{3n+1}$

Answer 1

The correct option is B $\frac{n}{n+1}$

For any integer $n\ge 1$ , let $p_n$ be the statement that

$\frac{1}{1.2}+\frac{1}{2.3}...\frac{1}{n+1}=\frac{n}{n+1}$ .

$\underset{–}{Basecase}$ : The statement $P_1$ says that

$\frac{1}{1.2}=\frac{1}{1+1}$,

This is true.

$\underset{–}{Inductivestep}$: Fix $k\ge 1$ , and suppose that $P_k$ holds, that is,

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

It remains to show that $P_{k+1}$ holds, that is,

$\frac{1}{1.2}+\frac{1}{2.3}...\frac{1}{(k+1)(k+2)}=\frac{k+1}{k+2}$ .
Since we know the sum of the series till the kth term we can write the LHS as
$\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}$
$=\frac{k^2+2k+1}{(k+1)(k+2)}$
$=\frac{{(k+1)}^2}{(k+1)(k+2)}$
$=\frac{k+1}{k+2}$ = RHS
Therefore $P_{k+1}$ holds

Thus, by the principle of mathematical induction, for all $n\ge 1$, $P_n$ holds.