Write an expression for the n th term of the sequence $\frac { 1 } { 6 } , - \frac { 1 } { 36 } , \frac { 1 } { 216 } , - \frac { 1 } { 1296 } , \ldots$ Assume that n begins with 1.
A) $a _ { n } = \frac { ( - 1 ) ^ { n } } { 6 ^ { n + 2 } }$
B) $a _ { n } = \frac { ( - 1 ) ^ { n + 1 } } { 6 ^ { n + 1 } }$
C) $a _ { n } = \frac { ( - 1 ) ^ { n } } { 6 ^ { n + 1 } }$
D) $a _ { n } = \frac { ( - 1 ) ^ { n + 1 } } { 6 ^ { n } }$
E) $a _ { n } = \frac { ( - 1 ) ^ { n } } { 6 ^ { n } }$

Question

Quizplus · Accepted Answer

Looking at the sequence, notice that the denominators are powers of 6, indicating that the expression involves 6 raised to a power. Also, notice that the signs alternate between positive and negative, indicating the use of (-1)^n or (-1)^(n+1). Since the first term is positive, and the negative terms begin with the second term, we choose (-1)^(n+1) instead of (-1)^n. Finally, since the term is the reciprocal of the denominator, we end up with the expression $\frac{( - 1 ) ^ { n + 1 }}{6^{n}}$ which matches choice D.

Write an Expression for the N Th Term of the Sequence