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 ????????? .

Well, well, well, let's solve this mathematical mystery together!

We know that in a geometric sequence with an initial term a and a common ratio r, each term can be found using the formula an = a * r^(n-1).

So, to find the terms of the sequence a1^2, a2^2, a3^2, ... let's start with the first term:

a1^2 = (a * r^(1-1))^2 = (a * r^0)^2 = a^2

Now, let's find the second term:

a2^2 = (a * r^(2-1))^2 = (a * r^1)^2 = a^2 * r^2

And the third term:

a3^2 = (a * r^(3-1))^2 = (a * r^2)^2 = a^2 * r^4

Hmm, do you see a pattern here? It looks like each term can be found by raising a^2 to a power of the increasing positive even integer!

So, the general formula for the terms of the sequence a1^2, a2^2, a3^2, ... is:

an^2 = a^2 * r^(2n-2)

And now, let's find the common ratio for this sequence. We can do that by taking the ratio of any two consecutive terms:

(a2^2) / (a1^2) = (a^2 * r^2) / (a^2) = r^2

Aha! The common ratio for the sequence a1^2, a2^2, a3^2, ... is indeed r^2.

So, we've cracked the code! The sequence a1^2, a2^2, a3^2, ... is a geometric sequence with a common ratio of r^2.

I hope this helps clarify things for you, and remember, math is no clowning matter!

To show that the sequence a1^2, a2^2, a3^2, ... is also a geometric sequence, we need to find its common ratio.

We know that the general term of the original geometric sequence a1, a2, a3, ... is given by an = a1 * r^(n-1).

To find the general term of the new sequence a1^2, a2^2, a3^2, ..., we simply need to square each term of the original sequence. So the general term of the new sequence is (a1 * r^(n-1))^2 = a1^2 * (r^(n-1))^2 = a1^2 * r^(2n-2).

Therefore, the general term of the new sequence is given by bn = a1^2 * r^(2n-2), where n = 1, 2, 3, ...

To find the common ratio of the new sequence, we can take the ratio of consecutive terms. Let's consider the ratio b(n+1)/bn:

b(n+1)/bn = (a1^2 * r^(2(n+1)-2))/(a1^2 * r^(2n-2))
= (a1^2 * r^(2n+2-2))/(a1^2 * r^(2n-2))
= r^(2n+2-2) / r^(2n-2)
= r^2 / 1
= r^2

Therefore, the common ratio of the new sequence a1^2, a2^2, a3^2, ... is r^2.

In conclusion, the sequence a1^2, a2^2, a3^2, ... is a geometric sequence with a common ratio of r^2.

To show that the sequence a₁², a₂², a₃², ... is also a geometric sequence, we need to find its common ratio.

We know that the general term of the original geometric sequence is given by:
an = a₁ * r^(n-1), where n = 1, 2, 3, ...

We want to find the general term of the sequence a₁², a₂², a₃², ..., so let's substitute an into the expression for an²:

an² = (a₁ * r^(n-1))²

Expanding the expression, we have:
an² = a₁² * (r^(n-1))²

Using the property (ab)² = a² * b², we can rewrite the expression as:
an² = a₁² * r^2(n-1)

Now, the question asks for the common ratio of the sequence a₁², a₂², a₃², ... which can be found by taking the ratio of consecutive terms. Let's calculate the ratio of a₂² to a₁²:

(a₂²) / (a₁²) = (a₁² * r^2(2-1)) / (a₁² * r^2(1-1))
= (a₁² * r^2) / a₁²

Simplifying the expression, we get:
(a₂²) / (a₁²) = r^2

So, the common ratio for the sequence a₁², a₂², a₃², ... is r².

In conclusion, the sequence a₁², a₂², a₃², ... is a geometric sequence with a common ratio of r².

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