Question

How do you evaluate `sum_(n=2)^6(2n-1)`?

Answer 1

`sum_(n=2)^6(2n-1)=35`

Explanation:

This appears to be an arithmetic series, but since there are only 5 terms, you might as well just write the terms out by moving the index of summation through from 2 to 6 and then add them :\

`therefore sum_(n=2)^6(2n-1)=3+5+7+9+11`

`=35`.

If you want to force the arithmetic series formula, then you may identify the common difference as 2, the first term as 3, and 5 terms in total in this convergent series to get

`S_n=n/2[2a+(n-1)d]`

`therefore S_6=5/2[(2)(3)+(5-1)(2)]`

`=35`.