A formula for the sum of all numbers divisible by 9 from 1 to 1000

A formula for the sum of all numbers divisible by 9 from 1 to 1000

9+18+27+..+9*111
=9(1+2+3+...+111)
=9(1+111)*111/2=55944

Find the sum of all numbers divisible by 9 between 200 and 300

207+216+.+297
=（207+297）*11/2=2772

-(radical 2-1) + 1 / (radical 3-radical 2)
=- radical 2 + 1 + (radical 3 + radical 2) / [(radical 3 + radical 2) (radical 3-radical 2)]
=- radical 2 + 1 + radical 3 + radical 2
=Root 3 + 1
I don't understand the second step,

This is called denominator rationalization, using the square difference formula: a * A-B * b = (a + b) * (a-b)
Use it the other way round
The numerator and denominator are all multiplied by (radical 3 + radical 2), so the denominator becomes (3-2) = 1
There are only molecules left
This method is very common