Question

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) $\bigcirc \quad a_n = \frac{(-1)^{2n+1}}{n^2}$ $\bigcirc \quad a_n = \frac{(-1)^n}{n^2}$ $\bigcirc \quad a_n = \frac{(-1)^{n+1}}{n^2}$ $\bigcirc \quad a_n = \frac{(-1)^{n^2}}{n^2}$

          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)
$\bigcirc \quad a_n = \frac{(-1)^{2n+1}}{n^2}$
$\bigcirc \quad a_n = \frac{(-1)^n}{n^2}$
$\bigcirc \quad a_n = \frac{(-1)^{n+1}}{n^2}$
$\bigcirc \quad a_n = \frac{(-1)^{n^2}}{n^2}$

Find a formula for the nth term of the sequence.
1, -(1)/(4), (1)/(9), -(1)/(16), (1)/(25) (reciprocals of squares with alternating signs)
◯ an = ((-1)^2n+1)/(n^2)
◯ an = ((-1)^n)/(n^2)
◯ an = ((-1)^n+1)/(n^2)
◯ an = ((-1)^n^2)/(n^2)

Added by Andrew C.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Khoi V

05/08/2020

Step 1

The given sequence is 1111, 49, 1625. Show more…

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Find a formula for the nth term of the sequence 1111 491625 (reciprocals of squares with alternating signs) =-12n+1 an n2 -1n n2 an=(-1yn+1 n2 -1n2 n2