pre-calculus urgent

Please help me here as well to clarify more my understanding.

Let a1, a2, a3, . . . be a geometric sequence with initial term a and common ratio r. Show that
a1^2, a2^2, a3^2, . . . is also a geometric sequence by finding its common ratio.

Given that the ratio for
a1, a2, a3, . . . is r, then an= a1r^n − 1, n = 1, 2, 3, . . .

an^2 = (a1 · ????????) ^2


= (a1^2) ( ??????????? )^n − 1

.
Therefore, the sequence
a12, a22, a32, . . . is a geometric with common ratio ????????? .

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asked by Alex
  1. a(n) = a(n-1)r
    [a(n)]^2 = [a(n-1)r]^2 = [a(n-1)]^2 r^2
    so common ratio is r^2 instead of r

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    posted by Damon

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