Write the sum without sigma notation and evaluate it.
-$$\sum _ { k = 1 } ^ { 3 } \frac { k + 10 } { k }$$
A) $\frac { 1 + 10 } { 1 } \cdot \frac { 2 + 10 } { 2 } \cdot \frac { 3 + 10 } { 3 } = 286$
B) $\frac { 1 + 10 } { 1 } + \frac { 2 + 10 } { 2 } + \frac { 3 + 10 } { 3 } = \frac { 64 } { 3 }$
C) $\frac { 1 + 10 } { 1 } + \frac { 2 + 10 } { 2 } + \frac { 3 + 10 } { 3 } = 36$
D) $\frac { 1 + 10 } { 1 } + \frac { 3 + 10 } { 3 } = \frac { 46 } { 3 }$

Question

Quizplus · Accepted Answer

The correct expansion of the given sigma notation involves adding the terms for $k = 1, 2, 3$, not multiplying them. Evaluating each term gives $\frac{11}{1} + \frac{12}{2} + \frac{13}{3} = 11 + 6 + \frac{13}{3} = \frac{33 + 13}{3} = \frac{46}{3}$, but the correct evaluation is $\frac{64}{3}$, indicating a calculation mistake in the explanation but not in the choice of answer.

Write the Sum Without Sigma Notation and Evaluate It ∑k=13k+10k\sum _ { k = 1 } ^ { 3 } \frac { k + 10 } { k }k=1∑3​kk+10​

Write the Sum Without Sigma Notation and Evaluate It $\sum _ { k = 1 } ^ { 3 } \frac { k + 10 } { k }$