Find a formula for the nth term of the sequence.
-$1 , - \frac { 1 } { 4 } , \frac { 1 } { 9 } , - \frac { 1 } { 16 } , \frac { 1 } { 25 }$ (reciprocals of squares with alternating signs)
A) $a _ { n } = \frac { ( - 1 ) ^ { n } } { n ^ { 2 } }$
B) $a _ { n } = \frac { ( - 1 ) ^ { 2 n + 1 } } { n ^ { 2 } }$
C) $a _ { n } = \frac { ( - 1 ) ^ { n + 1 } } { n ^ { 2 } }$
D) $a _ { n } = \frac { ( - 1 ) ^ { n ^ { 2 } } } { n ^ { 2 } }$

Question

Quizplus · Accepted Answer

The sequence is the reciprocals of the squares with alternating signs. The denominator is always the square of the index. The signs alternate between positive and negative, starting with a positive sign. Therefore, the formula for the nth term is $a_n = \frac{(-1)^{n+1}}{n^{2}}$. Choice C gives this formula.

Find a Formula for the Nth Term of the Sequence 1,−14,19,−116,1251 , - \frac { 1 } { 4 } , \frac { 1 } { 9 } , - \frac { 1 } { 16 } , \frac { 1 } { 25 }1,−41​,91​,−161​,251​

Find a Formula for the Nth Term of the Sequence $1 , - \frac { 1 } { 4 } , \frac { 1 } { 9 } , - \frac { 1 } { 16 } , \frac { 1 } { 25 }$