Given that the sum of the first n terms of the equal ratio sequence {an} is Sn = (2 ^ n) - 1, find the sum of the first n terms of the sequence {(an) ^ 2} and TN

Given that the sum of the first n terms of the equal ratio sequence {an} is Sn = (2 ^ n) - 1, find the sum of the first n terms of the sequence {(an) ^ 2} and TN

Because S1 = A1 = (2 ^ 1) - 1 = 1; also: SN = A1 * (1-Q ^ n) / (1-Q), and according to the question, q = 2
So, an = A1 * q ^ (n-1) = 2 ^ (n-1)
Then BN = {(an) ^ 2} = 2 ^ (2n-2)
In the sequence BN, B1 = 1, common ratio q = 4
So TN = B1 * (1-Q ^ n) / (1-Q) = (4 ^ n-1) / 3

The sum of the first n terms of the proportional sequence {an} is Sn if s2n = 3 (a1 + a3 +...) +Then the relationship between Sn and an is

When n = 1,
s2n=s2=3a1,
a2=s2-a1=2a1
q=2
a2^3=8
a2=2
a1=1
an=2^(n-1)
sn=2^n-1
=2x2^(n-1)-1
=2an -1

Let the sum of the first n terms of the equal ratio sequence {an} be Sn, if s2n = 4 (a1 + a3 + L + a (2n-1), a1a2a3 = 8, then A5=

Because s2n = 3 (a1 + a3 +...) +a2n－1),
Let n = 5
So S10 = 3 (a1 + a3 + A5 +... A9) = (a1 +. A10)
SO 2 (a1 + a3 + A5 +... A9) = (A2 + A4 + A6 +. A10)
q=2
Because a1a2a3 = 8
So A2 = 2
a1=1
So A5 = (2) ^ 4 = 16