Question

10 points for best answer!

Answer 1

`$\sum\limits_(n=1)^(15) (2n-1)`

When n = 1, 2n - 1 = 2(1) - 1 = 1
When n = 2, 2n - 1 = 2(2) - 1 = 3
When n = 3, 2n - 1 = 2(3) - 1 = 5
...

So: the total will be :

`= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29`

`= 225`


`$\sum\limits_(n=1)^(15) (2n-1) = 225`