Again, we have a sum of an arithmetic sequence. How many terms are in the sequence? Clearly each term in the sequence has the form \(4k -2\) (as evidenced by the last term). For which values of \(k\) though? To get 6, \(k = 2\text{.}\) To get \(4n-2\) take \(k = n\text{.}\) So to find the number of terms, we must count the number of integers in the range \(2,3,\ldots, n\text{.}\) This is \(n-1\text{.}\) (There are \(n\) numbers from 1 to \(n\text{,}\) so one less if we start with 2.)
Now reverse and add:
| \(S =\) | \(6\) | \(+\) | \(10\) | \(+ \cdots +\) | \(4n-6\) | \(+\) | \(4n-2\) |
| \(+ \quad S =\) | \(4n-2\) | \(+\) | \(4n-6\) | \(+ \cdots +\) | \(10\) | \(+\) | 6 |
| \(2S =\) | \(4n+4\) | \(+\) | \(4n+4\) | \(+ \cdots +\) | \(4n+4\) | \(+\) | \(4n+4\) |
Since there are \(n-1\) terms, we get