Question

Prove by induction that for every n ≥ 1 we have?

Answer 1

`n/(2n+1)+1/[(2(n+1)-1)(2(n+1)+1)]`
=`(n+1)/(2(n+1)+1)`
Which checks with the series for term (n+1)

Explanation:

Let `S_n` be our expression, i.e. `1/(1*3)+ ... + 1/[(2n-1)(2n+1)]`

First we must check that it fits for n=1:

Left side: `S_1=1/(1*3)=1/3`

Right side: `=1/(2*1+1)=1/3` Check

So if it is correct for `S_n`, we want to show that
`S_(n+1) = (n+1)/[2(n+1)+1)`

We have:
`S_(n+1)=S_n+1/[(2(n+1)-1)(2(n+1)+1)]`

`n/(2n+1)+1/[(2(n+1)-1)(2(n+1)+1)]`

=`n/(2n+1)+1/[(2n+1)(2n+3)]`

=`(n(2n+3)+1)/[(2n+1)(2n+3)]`= `(2n^2+3n+1)/[(2n+1)(2n+3)]`

=`((2n+1)(n+1))/[(2n+1)(2n+3)]`

=`(n+1)/(2n+3) =(n+1)/(2(n+1)+1)`

As this is the same as the right side of our series for term (n+1), the expression checks.