Question

Question 6 4 pts Write a formula for the general term (the nth term) of the geometric sequence. 4, 1, \frac{1}{4}, \frac{1}{16}, \frac{1}{64}, ... a_n = 4 \left(\frac{1}{4}\right)^{n-1} a_n = 4 \left(\frac{1}{4}\right)^{n+1} a_n = 4 \left(\frac{1}{4}\right)^n a_n = 4 \left(\frac{1}{16}\right)^{n-1}

          Question 6
4 pts
Write a formula for the general term (the nth term) of the geometric sequence.
4, 1, \frac{1}{4}, \frac{1}{16}, \frac{1}{64}, ...
a_n = 4 \left(\frac{1}{4}\right)^{n-1}
a_n = 4 \left(\frac{1}{4}\right)^{n+1}
a_n = 4 \left(\frac{1}{4}\right)^n
a_n = 4 \left(\frac{1}{16}\right)^{n-1}

Question 6
4 pts
Write a formula for the general term (the nth term) of the geometric sequence.
4, 1, (1)/(4), (1)/(16), (1)/(64), ...
an = 4 ((1)/(4))^n-1
an = 4 ((1)/(4))^n+1
an = 4 ((1)/(4))^n
an = 4 ((1)/(16))^n-1

Added by Lu-S J.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Mukesh Devi

Step 1

Identify the common ratio (r) between the terms: To find the common ratio, divide the second term by the first term, or the third term by the second term: r = 16/4 = 4 or 64/16 = 4 Show more…

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