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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)

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15 Nov 2011, 08:22
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Difficulty:

75% (hard)

Question Stats:

57% (03:41) correct 43% (02:59) wrong based on 224 sessions

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$$\frac{3(4^5+4^6+4^7+4^8+4^9+4^{10})}{(2^5+2^6+2^7+2^8+2^9+2^{10})} = ?$$

(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!
[Reveal] Spoiler: OA

Last edited by alinomoto on 15 Nov 2011, 11:49, edited 2 times in total.
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Re: Difficult Geometric series question. [#permalink]

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15 Nov 2011, 10:16
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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]

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15 Nov 2011, 11:30
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kostyan5 wrote:
I used mostly brute force, with some factoring, to solve this. I'd be interested in a more elegant solution.

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

$$=\frac{3*2^5*1365}{63} = \frac{2^5*1365}{21} = 2^5*65 = 2^5*(64+1) = 2^5*(2^6+1) = 2^{11}+2^5$$

Yes, that is the way I was going at it too, but I realised that there has to be an easier way to do this under 2 minutes.

$$\frac{3*(4^5+4^6+4^7+4^8+4^9+4^{10})}{(2^5+2^6+2^7+2^8+2^9+2^{10})}$$

just consider the top and bottom halves of the geometric progression.

S1 = $$(4^5+4^6+4^7+4^8+4^9+4^{10})$$

and S2 = $$(2^5+2^6+2^7+2^8+2^9+2^{10})$$

Solving for S1, since the common fraction in the GP is 4, multiply S1 by 4.

Thus, 4S1 = $$4*(4^5+4^6+4^7+4^8+4^9+4^{10}) = (4^6+4^7+4^8+4^9+4^{10}+4^{11})$$

Subtract 4S1 - S1 = 3S1 = $$(4^{11} - 4^5)$$

or S1 = $$\frac{(4^{11} - 4^5)}{3}$$

Similarly, for S2, multiply by common factor (2) and subtract:

2S2 - S2 = $$(2^{11} - 2^5)$$

Notice that 3S1 is the numerator cancels with the 3 of S1.

Thus we get $$\frac{3 *(4^{11} - 4^5)}{3} * \frac{1}{(2^{11} - 2^5)}$$.

Solving and factoring:
$$\frac{(4^{11} - 4^5)}{(2^{11} - 2^5)} = \frac{((2^2)^{11} - (2^2)^5)}{(2^{11} - 2^5)} = \frac{((2^{11})^2 - (2^5)^2)}{(2^{11} - 2^5)}$$

Now you see that the numerator takes the form $$a^2 - b^2 = (a+b) (a-b)$$

Solve and you get answer (C) $$2^{11} + 2^5$$
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Re: Difficult Geometric series question. [#permalink]

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15 Nov 2011, 14:53
Thanks. Neat trick.
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Re: Difficult Geometric series question. [#permalink]

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15 Nov 2011, 21:31
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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!

It is certainly not a GMAT-type question so I wouldn't worry about it.
Though, you can do it quickly using the formula Sum of GP = a(r^n - 1)/(r - 1)
a = first term, r = common ratio (which is 4 in the numerator) and n = number of terms

$$(4^5+4^6+4^7+4^8+4^9+4^{10}) = 4^5(4^6 - 1)/(4-1) = 4^5(4^6 - 1)/3$$

$$(2^5+2^6+2^7+2^8+2^9+2^{10}) = 2^5(2^6 - 1)/(2-1)$$

The required fraction becomes: $$\frac{4^5(4^6 - 1)}{2^5(2^6 - 1)}$$

= $$\frac{2^5(2^6 + 1)(2^6 - 1)}{(2^6 - 1)}$$ (as done above)
= $$2^{11} + 2^5$$
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16 Sep 2013, 04:50
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16 Sep 2013, 05:00
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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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2-99058.html
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26 Feb 2014, 21:30
Just want to confirm, is this a GMAT question??
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27 Feb 2014, 05:23
PareshGmat wrote:
Just want to confirm, is this a GMAT question??

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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30 Mar 2014, 09:09
@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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22 Jul 2015, 02:11
Hello from the GMAT Club BumpBot!

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31 Jul 2015, 19:53
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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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14 Sep 2016, 03:45
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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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] 14 Sep 2016, 03:45
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