## Calculus (3rd Edition)

Recall: $$\sum_{i=m}^{n}C=C(n-m+1)$$ (a) $$\sum_{i=1}^{5}9=9(5-1+1)=45$$ (b) $$\sum_{i=0}^{5} =4(5-0+1)=4\times6=24$$ (c) $$\sum_{k=2}^{4}k^{3}=2^{3}+3^{3}+4 {3}=8+27+64=99$$