Question

In Problems 1–22, use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers $n.$

$\frac{1}{1·2}+\frac{1}{2·3}+\frac{1}{3·4}+...+\frac{1}{n\left(n+1\right)}=\frac{n}{n+1}$

Short answer

The statement$\frac{1}{1·2}+\frac{1}{2·3}+\frac{1}{3·4}+...+\frac{1}{n\left(n+1\right)}=\frac{n}{n+1}$is true for all natural numbers.

Step-by-step solution

Step 1. Given information

The given statement is$\frac{1}{1·2}+\frac{1}{2·3}+\frac{1}{3·4}+...+\frac{1}{n\left(n+1\right)}=\frac{n}{n+1}.$We need to prove that the given statement is true for all natural numbers by using mathematical induction.

Step 2. Proof

Let's prove that the statement is true for $n=1.$

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

$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right..$

The statement is true for $n=1.$

Let's assume that the statement is true for $n=k.$

$\frac{1}{1·2}+\frac{1}{2·3}+\frac{1}{3·4}+...+\frac{1}{k\left(k+1\right)}=\frac{k}{k+1}.$

Let's prove that the statement is true for $n=k+1.$

$\frac{1}{1·2}+\frac{1}{2·3}+\frac{1}{3·4}+...+\frac{1}{k\left(k+1\right)}+\frac{1}{\left(k+1\right)\left(k+2\right)}.$

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

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

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

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

The statement is true for $n=k+1.$

Since the given statement is true for $n=1$and if the statement is true for some $n=k$and it is also true for the next natural number $n=k+1.$

So by the principle of mathematical induction, the statement is true for all natural numbers.