Did not find this in a previous thread.

2+2+(2^2)+(2^3)+(2^4)+(2^5)+(2^6)+(2^7)+(2^8)

2^9

2^10

2^16

2^35

2^37

Is there a way to do this problem without calculating the powers of 2? Thanks.

2+2+(2^2)+(2^3)+(2^4)+(2^5)+(2^6)+(2^7)+(2^8)

after the first two terms i.e (2^2)+(2^3)+(2^4)+(2^5)+(2^6)+(2^7)+(2^8) is a GP series with first term as 2^2 and ratio as 2.
Using the formula for sum of GP series, for this part, the original equation becomes
2^2 + 2^2(1-2^7)/(1-2)
= 2^2 + 2^2(127)
= 2^2(1 + 127)
= 2^2 * 2^7
=2^9

there are several ways.

(1) PATTERN RECOGNITION
it should be clear that there's nothing special about 2^8 as an ending point; in other words, they just cut the sequence off at a random point. therefore, if we investigate smaller "versions" of the sequence, we should be able to detect a pattern.
let's look:
first term = 2
sum of first 2 terms = 4
sum of first 3 terms = 8
sum of first 4 terms = 16
ok, it's clear what's going on: each new term doubles the sum. if you see a pattern this clear, it doesn't batter whether you understand WHY the pattern exists; just continue it.
so, i want the sum of nine terms, so i'll just double the sum five more times:
32, 64, 128, 256, 512.
this is choice (a).
this is a general rule, by the way: IF SOMETHING CONTAINS MORE THAN 4-5 IDENTICAL STEPS, YOU SHOULD BE ABLE TO EXTRACT A PATTERN FROM LOOKING AT SIMILAR EXAMPLES WITH FEWER STEPS.

(2) ALGEBRA WITH EXPONENTS ("textbook method")
the first two terms are 2 + 2. this is 2(2), or 2^2.
now, using this combined term as the "first term", the first two terms are 2^2 + 2^2. this is 2(2^2), or (2^1)(2^2), or 2^3.
now, using this combined term as the "first term", the first two terms are 2^3 + 2^3. this is 2(2^3), or (2^1)(2^3), or 2^4.
you can see that this will keep happening, so it will continue all the way up to 2^8 + 2^8, which is 2(2^8) = (2^1)(2^8) = 2^9.

(3) ESTIMATE
these answer choices are ridiculously far apart, so you should be able to estimate the answer.
memorize some select powers of 2. notably, 2^10 = 1024, which is "about 1000". 2^9 = 512, which is "about 500". and of course you should know all the smaller ones (2^6 and below) by heart.
thus we have 2^8 is about 250, and the other terms are 128, 64, 32, 16, 8, 4, 2, 2.
looking at these numbers, i'd make a ROUGH ESTIMATE WITHIN A FEW SECONDS:
250 is 250.
128 is ~130.
64 and 32 together are ~100.
the others look like thirty or so together.
so, 250 + 130 + 100 + 30 = 510.
the only answer choice within shouting range is (a); the others are absurdly huge.

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even if you have no idea how to do anything else, you should still be able to do out the arithmetic within the two-minute time limit.
it won't be fun, but you should be able to do it. if you can't, then the reason is probably "you stared at the problem for too long, and didn't get started when you should have".

yes ,the shortest method on this planet to solve the above question.

This formula may be of use:2^1+2^2+........+2^n =[2^(n+1)] - 2, where n equals to number of terms.

question is: 2+2+(2^2)+(2^3)+(2^4)+(2^5)+(2^6)+(2^7)+(2^8)

this can be written as : 2+(2^1)+(2^2)+(2^3)+(2^4)+(2^5)+(2^6)+(2^7)+(2^8)

Therefore, 2+[2^(8+1)] - 2 = 2^9 (answer)

Thanks rohan_sisodiya for the formulaic approach to do this problem. If you do know the formula, then it will get to the answer quickly. Otherwise, any of Ron's approaches are very helpful!

_________________
Ben Ku
Instructor
ManhattanGMAT

In this case, the numbers were pretty small.
so a little bit grouping helped me do the arithmetic directly.
= 2^2 [1+1+2+(2^2)+(2^3)+(2^4)+(2^5)+(2^6)]
= 4*[1+1+2+4+8+16+32+64] = 4*128
= 512

The exponent formula in the previous post and ron's pattern recognition technique have gone into my 'takeaways' log. :)

If you look at the pattern, you can deduce the answer as 2^9 without doing any calculations.

2+2+(2^2)+(2^3)+(2^4)+(2^5)+(2^6)+(2^7)+(2^8) =
(2^2)+(2^2)+(2^3)+(2^4)+(2^5)+(2^6)+(2^7)+(2^8) =
(2^3)+(2^3)+(2^4)+(2^5)+(2^6)+(2^7)+(2^8) =
(2^4)+(2^4)+(2^5)+(2^6)+(2^7)+(2^8)

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(2^8) + (2^8) = 2^9

this is already listed as method #2 in my post, four posts above yours.
please try not to duplicate existing content in these threads; thanks.