Provide an appropriate response.
-Which formula is not equivalent to the other two?
I. $\sum _ { k = 3 } ^ { 5 } \frac { ( - 1 ) ^ { \mathrm { k } + 1 } } { \mathrm { k } + 1 }$
II. $\sum _ { k = 1 } ^ { 3 } \frac { ( - 1 ) ^ { k + 3 } } { k + 3 }$
III. $\sum _ { k = 2 } ^ { 4 } \frac { ( - 1 ) ^ { k - 1 } } { k - 1 }$
A) II
B) All are equivalent.
C) III
D) $\mathrm { I }$

Question

Quizplus · Accepted Answer

To check which formula is not equivalent to the other two, we expand each of the summations and simplify:
I. $\sum _ { k = 3 } ^ { 5 } \frac { ( - 1 ) ^ { \mathrm { k } + 1 } } { \mathrm { k } + 1 } = \frac{-1}{4} + \frac{1}{5} - \frac{1}{6} = \frac{-11}{60}$
II. $\sum _ { k = 1 } ^ { 3 } \frac { ( - 1 ) ^ { k + 3 } } { k + 3 } = \frac{1}{4} - \frac{1}{5} + \frac{1}{6} = \frac{1}{120}$
III. $\sum _ { k = 2 } ^ { 4 } \frac { ( - 1 ) ^ { k - 1 } } { k - 1 } = \frac{-1}{1} + \frac{1}{2} - \frac{1}{3} = \frac{-1}{3}$
We can see that the sum in formula II is not equivalent to the other two, as it has a different value. Therefore, the answer is C) III.

Provide an Appropriate Response ∑k=35(−1)k+1k+1\sum _ { k = 3 } ^ { 5 } \frac { ( - 1 ) ^ { \mathrm { k } + 1 } } { \mathrm { k } + 1 }∑k=35​k+1(−1)k+1​

Provide an Appropriate Response $\sum _ { k = 3 } ^ { 5 } \frac { ( - 1 ) ^ { \mathrm { k } + 1 } } { \mathrm { k } + 1 }$