Chapter 5 - Section 5.2 - Sigma Notation and Limits of Finite Sums - Exercises - Page 299: 29

a) $21$ b) $3500$ c) $2620$

Use formula: $\Sigma_{k=1}^n k=\dfrac{n(n+1)}{2}$ and $\Sigma_{k=1}^n k^2=\dfrac{n(n+1)(n+2)}{6}$ a) Here, we have $\Sigma_{k=1}^{7} (3)=(3)(7)=21$ b) Here, we have $\Sigma_{k=1}^{500}(7) =(500)(7)=3500$ c) Here, we have $\Sigma_{k=3}^{264} 10=\Sigma_{j=1}^{262} 10=(262)(10)=2620$