Question

Prove by induction:
$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....$ to $n$ terms$=\frac{n}{n+1}$

Answer 1

The statement to be proved is:

$P\left(n\right):\frac{1}{1.2}+\frac{1}{2.3}+...$ to $n$ terms $=\frac{n}{n+1}$

Step 1: Prove that the statement is true for $n=1$

$P\left(1\right):\frac{1}{1.2}=\frac{1}{1+1}$

$P\left(1\right):\frac{1}{2}=\frac{1}{2}$

Hence, the statement is true for $n=1$

Step 2: Assume that the statement is true for $n=k$

Let us assume that the below statement is true:

$P\left(k\right):\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{k(k+1)}=\frac{k}{k+1}$

Step 3: Prove that the statement is true for $n=k+1$

We need to prove that:

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

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

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

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

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

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

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

$=\frac{k+1}{k+2}$

$=RHS$

Therefore, the statement is true for all $n$ by principle of mathematical induction.