Chapter 4 - Integration - 4.4 The Definition Of Area As A Limit; Sigma Notation - Exercises Set 4.4 - Page 298: 12

$$35450$$

Given $$\sum_{k=1}^{100}(7k+1)$$ Since $$\sum_{k=1}^{n}k=\frac{n(n+1)}{2},\ \ \sum_{k=1}^{n}1=n $$ Then \begin{align*} \sum_{k=1}^{100}(7k+1)&=7\sum_{k=1}^{100}k+\sum_{k=1}^{100}1\\ &=\frac{(7)(100)(101)}{2}+100 \\ &=35450 \end{align*}