Question

write the sum without sigma notation. then evaluate.

$$sum _ { k = 1 } ^ { 3 } ( - 1 ) ^ { k + 1 } cos k pi$$

write the sum without sigma notation.

$$( - 1 ) ^ { 1 + 1 } cos ( pi ) + ( - 1 ) ^ { 2 + 1 } cos ( 2 pi ) + ( - 1 ) ^ { 3 + 1 } cos ( 3 pi )$$
(do not simplify. do not evaluate.)

evaluate the sum.

$$sum _ { k = 1 } ^ { 3 } ( - 1 ) ^ { k + 1 } cos k pi = square$$
(simplify your answer. type an exact answer, using radicals as needed )

Explanation:

Step1: Calculate each term

• When \(k = 1\):

\((-1)^{1 + 1}\cos(1\pi)=(-1)^{2}\cos(\pi)=1\times(- 1)=-1\)

• When \(k = 2\):

\((-1)^{2 + 1}\cos(2\pi)=(-1)^{3}\cos(2\pi)=(-1)\times1=-1\)

• When \(k = 3\):

\((-1)^{3 + 1}\cos(3\pi)=(-1)^{4}\cos(3\pi)=1\times(-1)=-1\)

Step2: Sum up the terms

\(\sum_{k = 1}^{3}(-1)^{k + 1}\cos(k\pi)=(-1)+(-1)+(-1)=-3\)

Answer:

\(-3\)