2+3+5+6+8+9+....+(3n-1)+(3n)

just add up the two series

2+5+8+...+(3n-1)
3+6+9+...+3n

those are easy, right?

The series you provided is an arithmetic series in which each term is obtained by adding 3 to the previous term. To find the sum of the series up to the nth term, we can use the formula for the sum of an arithmetic series.

The formula for the sum of an arithmetic series is given by:

Sn = (n/2)(2a + (n-1)d)

Where:
- Sn represents the sum of the series.
- n is the number of terms in the series.
- a is the first term of the series.
- d is the common difference between consecutive terms.

In the given series, the first term (a) is 2, and the common difference (d) is 3. Therefore, we need to determine the nth term of the series.

The nth term of an arithmetic series is given by the formula:

an = a + (n-1)d

In this case, the nth term (an) is (3n) because the terms in the series are increasing by 3. We want to find the sum of the series up to (3n).

Now we can substitute these values into the formula for the sum of an arithmetic series:

Sn = (n/2)(2a + (n-1)d)
= (n/2)(2(2) + (n-1)(3))
= (n/2)(4 + 3n - 3)
= (n/2)(3n + 1)

Therefore, the sum of the series 2+3+5+6+8+9+....+(3n-1)+(3n) is (n/2)(3n + 1).