Question

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

$\frac{1}{1·3}+\frac{1}{3·5}+\frac{1}{5·7}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}=\frac{n}{2n+1}$

Short answer

The statement$\frac{1}{1·3}+\frac{1}{3·5}+\frac{1}{5·7}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}=\frac{n}{2n+1}$is true for all natural numbers.

Step-by-step solution

Step 1. Given information

The given statement is$\frac{1}{1·3}+\frac{1}{3·5}+\frac{1}{5·7}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}=\frac{n}{2n+1}.$We need to prove that the given statement is true for all natural numbers by using mathematical induction.

Step 2. Proof

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

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

$\frac{1}{1·3}=\frac{1}{2+1}.$

$\frac{1}{3}=\frac{1}{3}.$

So, the statement is true for$n=1.$

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

$\frac{1}{1·3}+\frac{1}{3·5}+\frac{1}{5·7}+...+\frac{1}{\left(2k-1\right)\left(2k+1\right)}=\frac{k}{2k+1}.$

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

$\frac{1}{1·3}+\frac{1}{3·5}+\frac{1}{5·7}+...+\frac{1}{\left(2k-1\right)\left(2k+1\right)}\frac{1}{\left\{2\left(k+1\right)-1\right\}\left\{2\left(k+1\right)+1\right\}}.$

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

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

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

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

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

$=\frac{k+1}{2k+3}.$

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

So, 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.