## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$1110$
Recall the formulas: $$A) \displaystyle \sum_{k=1}^{n} k =\dfrac{n(n+1)}{2} ;\\ B) \sum_{k=1}^{n} c=c+c+c+...+c=cn ; \\C) \sum_{k=1}^{n} (k-c) = \sum_{k=1}^{n} k- \sum_{k=1}^{n} c ; \ \\ D) \sum_{k=1}^{n} (c k)=c \sum_{k=1}^{n} (k)$$ Use formula $(C)$ to obtain: $\displaystyle \sum_{k=1}^{20} (5k+3) = \sum_{k=1}^{20} (5k) + \sum_{k=1}^{20} (3)$ Use formula $(D)$ to obtain: $\displaystyle \sum_{k=1}^{20} (5k)+\sum_{k=1}^{20} (3)=5 \sum_{k=1}^{20} (k) +\sum_{k=1}^{20} (3)$ Finally, apply formulas $(A)$ and $(B)$ to obtain: $5 \displaystyle \sum_{k=1}^{20} (k) + \sum_{k=1}^{20} (3) = (5) [\dfrac{20(20+1)}{2}] +(20)(3) \\=1050+60 \\=1110$