Write the sum without sigma notation and evaluate it.
-$$\sum _ { k = 1 } ^ { 3 } 2 ^ { k } \sin \frac { \pi } { 6 }$$
A) $2 + 4 + 8 = 14$
B) $2 \sin \frac { \pi } { 6 } + 4 \sin \frac { \pi } { 6 } + 8 \sin \frac { \pi } { 6 } = 7 \sqrt { 3 }$
C) $2 \sin \frac { \pi } { 6 } + 8 \sin \frac { \pi } { 6 } = 5$
D) $2 \sin \frac { \pi } { 6 } + 4 \sin \frac { \pi } { 6 } + 8 \sin \frac { \pi } { 6 } = 7$

Question

Quizplus · Accepted Answer

The sum without sigma notation is $2^1\sin\frac{\pi}{6} + 2^2\sin\frac{\pi}{6} + 2^3\sin\frac{\pi}{6}$. Since $\sin\frac{\pi}{6} = \frac{1}{2}$, the sum evaluates to $1 + 2 + 4 = 7$.

Write the Sum Without Sigma Notation and Evaluate It ∑k=132ksin⁡π6\sum _ { k = 1 } ^ { 3 } 2 ^ { k } \sin \frac { \pi } { 6 }k=1∑3​2ksin6π​

Write the Sum Without Sigma Notation and Evaluate It $\sum _ { k = 1 } ^ { 3 } 2 ^ { k } \sin \frac { \pi } { 6 }$