how's it E?
please explain.

I'll post a solution in 2 days.
In the meantime, see you do with the question.

Cheers,
Brent

The first term of the sequence is given as 64 = 2^6
From the formula,
2nd term = \(2^2 * 2^6\)
3rd term = \(2^3 * 2^2 * 2^6\)
4th term = \(2^4 * 2^3 * 2^2 * 2^6\)

There is a pattern with 2^6 constant for every term and the power of 2 is raised consecutively from \(2^2 to 2^n\)

Hence, \(2^1^1 = 2^6 * 2^2 * 2^3 ...........*2^1^1\)
similarly \(2^8 = 2^6 * 2^2 * 2^3.........*2^8\)

when finding the ratio everything cancels leaving only \(2^9 * 2^10 * 2^1^1 = 2^3^0\)

*kudos for all correct solutions[/quote]

Let's list a few terms and look for a pattern

term1 = 64 = 2^6
term2 = (2^6)(2^2)
term3 = (2^6)(2^2)(2^3)
term4 = (2^6)(2^2)(2^3)(2^4)
.
.
.
term8 = (2^6)(2^2)(2^3)(2^4)(2^5)(2^6)(2^7)(2^8)
.
.
.
term11 = (2^6)(2^2)(2^3)(2^4)(2^5)(2^6)(2^7)(2^8)(2^9)(2^10)(2^11)

So, term11/term8 = (2^6)(2^2)(2^3)(2^4)(2^5)(2^6)(2^7)(2^8)(2^9)(2^10)(2^11)/(2^6)(2^2)(2^3)(2^4)(2^5)(2^6)(2^7)(2^8)
= (2^9)(2^10)(2^11)
= 2^30

Answer: E

Cheers,
Brent

Let's list a few terms and look for a pattern

term1 = 64 = 2^8
term2 = (2^8)(2^2)
term3 = (2^8)(2^2)(2^3)
term4 = (2^8)(2^2)(2^3)(2^4)
.
.
.
term8 = (2^8)(2^2)(2^3)(2^4)(2^5)(2^6)(2^7)(2^8)
.
.
.
term11 = (2^8)(2^2)(2^3)(2^4)(2^5)(2^6)(2^7)(2^8)(2^9)(2^10)(2^11)

So, term11/term8 = (2^8)(2^2)(2^3)(2^4)(2^5)(2^6)(2^7)(2^8)(2^9)(2^10)(2^11)/(2^8)(2^2)(2^3)(2^4)(2^5)(2^6)(2^7)(2^8)
= (2^9)(2^10)(2^11)
= 2^30

Answer: E

Cheers,
Brent[/quote]

64=2^8 or 2^6

Good catch!
Rookie mistake on my part.
64 = 2^6 (not 2^8)

That said, the two expressions with 2^8 (or 2^6) cancel each other out. So, fortunately, the solution still worked out

Cheers and thanks for the heads up!
I have edited my answer accordingly.

Cheers,
Brent

\(\frac{t_{11}}{t_8} = \frac{2^{11}t_{10}}{t_8} = \frac{(2^{11}*2^{10})t_9}{t_8} = \frac{(2^{11}*2^{10}* 2^9)t_8}{t_8} = 2^{11}*2^{10}*2^9 = 2^{11 + 10 + 9} = 2^{30}\)

Hence, Answer is E

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