Write the sum without sigma notation and evaluate it.
-$$\sum _ { k = 1 } ^ { 3 } \frac { k + 8 } { k }$$
A) $\frac { 1 + 8 } { 1 } + \frac { 2 + 8 } { 2 } + \frac { 3 + 8 } { 3 } = \frac { 53 } { 3 }$
B) $\frac { 1 + 8 } { 1 } \cdot \frac { 2 + 8 } { 2 } \cdot \frac { 3 + 8 } { 3 } = 165$
C) $\frac { 1 + 8 } { 1 } + \frac { 3 + 8 } { 3 } = \frac { 38 } { 3 }$
D) $\frac { 1 + 8 } { 1 } + \frac { 2 + 8 } { 2 } + \frac { 3 + 8 } { 3 } = 30$

Question

Quizplus · Accepted Answer

We simply substitute the values of k and add the resulting fractions:
$$\frac{1+8}{1}+\frac{2+8}{2}+\frac{3+8}{3}=\frac{9}{1}+\frac{10}{2}+\frac{11}{3}=\frac{27+20+33}{6}=\frac{80}{6}=\frac{40}{3}=\boxed{\frac{53}{3}}$$

Write the Sum Without Sigma Notation and Evaluate It ∑k=13k+8k\sum _ { k = 1 } ^ { 3 } \frac { k + 8 } { k }k=1∑3​kk+8​

Write the Sum Without Sigma Notation and Evaluate It $\sum _ { k = 1 } ^ { 3 } \frac { k + 8 } { k }$