Question

A geometric sequence is defined by the recursive formula t1 = 64, tn = tn - 1 2 , where n ∈N and n > 1. The sequence is

Answer 1

A(n)=64(2^(n-1))

......................  oh you want to change is so n>1....okay...

64=a(2^n)  since the first term is 64...

64=2a so a=32 then...

a(n)=32(2^n)

Answer 2

Answer with explanation:

⇒Formula defining Geometric Sequence and the sequence is

   Let a be the first term and r be the common difference.

$t_{1}=64\\\\t_{n}=(t_{n-1})^2\\\\ar^{n-1}=(ar^{n-2})^2\\\\a_{n}=ar^{n-1}\\\\ar^{n-1}=a^2r^{2n-2}\\\\a^{2-1}r^{2n-2-n+1}=1\\\\a*r^{n-1}=1\\\\64*r^{n-1}=1\\\\r^{n-1}=\frac{1}{64}\\\\r^{n-1}=[\frac{1}{4}]^3\\\\r=\frac{1}{4}\\\\ \text{So,the sequence is}\\\\64,64*\frac{1}{4},64*[\frac{1}{4}]^2,64*[\frac{1}{4}]^3,64*[\frac{1}{4}]^4,.......\\\\=64, 16,4,1,\frac{1}{4},\frac{1}{16},.....$