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) $a _ { n } = \frac { ( - 1 ) ^ { n ^ { 2 } } } { n ^ { 2 } }$
B) $a _ { n } = \frac { ( - 1 ) ^ { n } } { n ^ { 2 } }$
C) $a _ { n } = \frac { ( - 1 ) ^ { 2 n + 1 } } { n ^ { 2 } }$
D) $a _ { n } = \frac { ( - 1 ) ^ { n + 1 } } { n ^ { 2 } }$

Question

Quizplus · Accepted Answer

The sequence is formed by taking the reciprocals of perfect squares and changing the sign after each term. The general formula for the nth term is given by $a_n=\frac{(-1)^{n+1}}{n^2}$, as it alternates the sign after each term and takes the reciprocal of the square of the value of n. Choice D matches this formula, so it is the best choice.

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 $1 , - \frac { 1 } { 4 } , \frac { 1 } { 9 } , - \frac { 1 } { 16 } , \frac { 1 } { 25 }$