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Re: Difficult Geometric series question. [#permalink]
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I used mostly brute force, with some factoring, to solve this. I'd be interested in a more elegant solution.

\(\frac{3*(4^5+4^6+4^7+4^8+4^9+4^{10})}{(2^5+2^6+2^7+2^8+2^9+2^{10})} = \frac{3*4^5(1+4^1+4^2+4^3+4^4+4^5)}{2^5*(1+2^1+2^2+2^3+2^4+2^5)} = \frac{3*2^5*2^5(1+4+16+64+256+1024)}{2^5*(1+2+4+8+16+32)} =\)

\(=\frac{3*2^5*1365}{63} = \frac{2^5*1365}{21} = 2^5*65 = 2^5*(64+1) = 2^5*(2^6+1) = 2^{11}+2^5\)
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Re: Difficult Geometric series question. [#permalink]
Thanks. Neat trick.
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Re: 3(4^5+4^6+4^7+4^8+4^9+4^10)/(2^5+2^6+2^7+2^8+2^9+2^10) [#permalink]
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alinomoto wrote:
\(\frac{3(4^5+4^6+4^7+4^8+4^9+4^{10})}{(2^5+2^6+2^7+2^8+2^9+2^{10})} = ?\)

(A) \(2^{10} + 2^5\)
(B) \(2^{10} + 2^6\)
(C) \(2^{11} + 2^5\)
(D) \(2^{11} + 2^6\)
(E) \(2^{16}\)


I was stuck at this question for a loooong time and I had to check the solution to see how it was done. Got it from the GMAT Quantum site. I haven't come across solving for sum of GP like this before in any of the Manhattan guides or any other GMAT question. Damn!


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Re: 3(4^5+4^6+4^7+4^8+4^9+4^10)/(2^5+2^6+2^7+2^8+2^9+2^10) [#permalink]
Just want to confirm, is this a GMAT question??
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Re: 3(4^5+4^6+4^7+4^8+4^9+4^10)/(2^5+2^6+2^7+2^8+2^9+2^10) [#permalink]
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PareshGmat wrote:
Just want to confirm, is this a GMAT question??


Please read Karishma's response:
VeritasPrepKarishma wrote:
alinomoto wrote:
\(\frac{3(4^5+4^6+4^7+4^8+4^9+4^{10})}{(2^5+2^6+2^7+2^8+2^9+2^{10})} = ?\)

(A) \(2^{10} + 2^5\)
(B) \(2^{10} + 2^6\)
(C) \(2^{11} + 2^5\)
(D) \(2^{11} + 2^6\)
(E) \(2^{16}\)


It is certainly not a GMAT-type question so I wouldn't worry about it.
...
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Re: 3(4^5+4^6+4^7+4^8+4^9+4^10)/(2^5+2^6+2^7+2^8+2^9+2^10) [#permalink]
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@PareshGMAT

I wrote this question and I want to assure everyone that this concept has been tested on the GMAT. However, one will encounter this question only if scoring close to Q50/Q51. I would recommend understanding the approach that @alinomoto took by multiplying the geometric series by the common factor and subtracting it. This same approach is used to obtain a general expression for the sum of the terms in a geometric sequence. One could memorize the formula for a geometric progression, but I personally think it is better to understand the underlying approach that is used to arrive at the expression.

Over the last four or five years, the GMAT test writers have had to introduce newer concepts to distinguish between students scoring at the upper percentile on the GMAT. The GMAT quant scores have been steadily creeping up over the last decade or so, the fact that Q51 is now 98% as opposed to 99% reflects this. The introduction of summation of a geometric sequence is an example of such a topic.

Cheers,
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3(4^5+4^6+4^7+4^8+4^9+4^10)/(2^5+2^6+2^7+2^8+2^9+2^10) [#permalink]
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I think this can be done without resort to special formulas if we recognize the \(21\)s in the numerator.

Specifically, once you factor and reduce down to \(\frac{3∗2^5∗(1+2^2+2^4+2^6+2^8+2^{10})}{1+2+4+8+16+32}\), you can add up the denominator to get 63, but adding the numerator is awful.

Instead, divide out the \(3\) to get \(\frac{2^5∗(1+2^2+2^4+2^6+2^8+2^{10})}{21}\).

There are no \(21\)s in the answer choices. So start adding the numerator and note that \(1+4+16 = 21\). Setting that aside, we can take the rest of the terms (\(2^6+2^8+2^{10}\)) and factor out \(2^6\). These terms also turn into \(1+4+16 = 21\). So we actually have \(\frac{2^5∗(21+2^6*(21))}{21} = 2^5∗(1+2^6) = 2^5+2^{11}\).
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Re: 3(4^5+4^6+4^7+4^8+4^9+4^10)/(2^5+2^6+2^7+2^8+2^9+2^10) [#permalink]
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Re: 3(4^5+4^6+4^7+4^8+4^9+4^10)/(2^5+2^6+2^7+2^8+2^9+2^10) [#permalink]
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