Question

How do you find the sum of the first 68 terms of the series `2+9+16+23+30+...`?

Answer 1

`16,082`

Explanation:

We have: `2 + 9 + 16 + 23 + 30 + ...`

This is an arithmetic series, so let's determine the common difference `d`:

`=> d = u_(2) - u_(1)`

`=> d = 9 - 2`

`=> d = 7`

Then, let's use the formula for the sum of `n` terms of an arithmetic sequence:

`=> S_(n) = (n) / (2) (2 u_(1) + (n - 1) d)`

Then, for the first `68` terms of the series:

`=> S_(68) = (68) / (2) (2 cdot 2 + (68 - 1) 7)`

`=> S_(68) = 34 (4 + 67 cdot 7)`

`=> S_(68) = 34 (4 + 469)`

`=> S_(68) = 34 (473)`

`=> S_(68) = 16,082`

Answer 2

Use the formulas `a_n=a_1+(n-1)d` and `S_n=(n(a_1+a_n))/2`.

Explanation:

This is an arithmetic series. The first term is `a_1=2`.
The common difference between each term `d= 7` because you add 7 to each term to get the next term. The number of terms is `n=68`.

Using the formula `a_n=a_1+(n-1)d`, calculate the 68th term.

`a_68 = 2+(68-1)7 = 471`

Next, use the formula `S_n=(n(a_1+a_n))/2` to find the sum of the first 68 terms.

`S_68= (68(2+471))/2 = 16082`