Answer
$1110$
Work Step by Step
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:
$5 \displaystyle \sum_{k=1}^{20} (k) + \sum_{k=1}^{20} (3) = (5) [\dfrac{20(20+1)}{2}] +(20)(3) \\=1050+60 \\=1110$
