What is the sum of the arithmetic series from (5 to n - 1) Sigma (3n - 2)

Solution:

The statement of the \(\sum_{5}^{n-1}3n-2\) problem is written as:

\(\sum_{5}^{n-1}3n-2\)

The series can be written as follows:

T(1), T(2), T(3), …….. T(n - 1)

T(1) = 3 × 5 - 2 = 13

T(2) = 3 × 6 - 2 = 16

T(3) = 3 × 7 - 2 = 19

T(4) = 3 × 8 - 2 = 22

T(n-1) = 3 ×(n-1) - 2 = 3n-5

The series is 13, 16, 19. 22, 25, …………….., 3n - 5

The first term a = 13 and common difference d = 3.

The number of terms is (n-5).

The last term l = is 3n-5.

The sum of n terms of an arithmetic series, when the first and the last terms are known is given by :

S_n = (n/2)[a+l]

Therefore the sum of the arithmetic series is :

S_{n - 5} = (n - 5)/2 [13 + 3n -5]

= (n - 5)(3n + 8)/2

---

What is the sum of the arithmetic series from (5 to n - 1) Sigma (3n - 2)

Summary:

The sum of the arithmetic series from (5 to n - 1) Sigma (3n - 2) is = (n - 5)(3n + 8)/2