# Calculate summation  `1/(1cdot3cdot5)+1/(3cdot5cdot7)+cdots+1/((2n-1)(2n+1)(2n+3))`

When you have summation like this, the bet thing is to calculate what sum for first few values of `n`  and then try to come up with some formula which connects those numbers.

For `n=1` `3/45`

`n=2`   `8/105`

`n=3`   `15/189`

`n=4`   `24/297`

Now there is no easy way to come up with formula. What you can do is to look separately numerator and denominator.

For numerator the formula is: `n(n+2)`

For denominator the formula is: `3(2n+1)(2n+3)`

So formula for our sum is `(n(n+2))/(3(2n+1)(2n+3))`

Now to be sure that our formula is valid for all `n in NN` we need to check this up by using mathematical induction.

1)

We know that formula is true for `n=1` because we constructed it to be valid for `n=1,2,3,4.`

2)

Let's assume that formula is valid for all integers greater or equal to `n.` i.e.

`1/(1cdot3cdot5)+1/(3cdot5cdot7)+cdots+1/((2n-1)(2n+1)(2n+3))=(n(n+2))/(3(2n+1)(2n+3))`

3)

Check formula for `n+1`

`1/(1cdot3cdot5)+1/(3cdot5cdot7)+cdots+1/((2n-1)(2n+1)(2n+3))+1/((2n+1)(2n+3)(2n+5))=`

`(n(n+2))/(3(2n+1)(2n+3))+1/((2n+1)(2n+3)(2n+5))=`

`(n(n+2)(2n+5)+3)/(3(2n+1)(2n+3)(2n+5))=(2n^3+5n^2+4n^2+10n+3)/(3(8n^3+36n^2+46n+15))=`

`(n^2+4n+3)/(3(4n^2+16n+15))=((n+1)(n+3))/(3(2(n+1)+1)(2(n+1)+3))`

Which proves that our formula is true for all natural numbers.

Approved by eNotes Editorial Team