Question 452857
The series is arithmetic with common difference -8. We can determine that the nth term is given by *[tex a_n = 50 - 8n] and express the sum of the first eight terms in sigma notation:


*[tex \LARGE \sum_{n=1}^8 a_n = \sum_{i=1}^8 50 - 8n]


There are multiple ways to evaluate the sum without using a calculator. Perhaps the easiest way is to manipulate the expression inside the sum:


*[tex \LARGE \sum_{n=1}^8 50 - 8n = \sum_{n=1}^8 50 - 8\sum_{n=1}^8 n]


The sum *[tex \sum_{n=1}^8 50] is simply 50 added eight times, or 400. The sum *[tex \sum_{n=1}^8 n] is equal to *[tex \frac{8(9)}{2} = 36]. Hence,


*[tex \LARGE \sum_{n=1}^8 50 - 8\sum_{n=1}^8 n = 400 - 8(36) = 112]