is there a quicker more efficient way to solve this problem without actually adding?
2+2+2^2+2^3+2^4+...+2^8
answer choices:
2^9
2^10
2^35
2^37
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exponent question on gmatprep
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truplayer256
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2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8
4=2^2
2(2^2)+2^3....+2^8
2(2^3)+2^4....+2^8
If you start to recognize the pattern in this question, it's clear that the answer is 2(2^8) or 2^9.
4=2^2
2(2^2)+2^3....+2^8
2(2^3)+2^4....+2^8
If you start to recognize the pattern in this question, it's clear that the answer is 2(2^8) or 2^9.
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PussInBoots
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Use simple logic: 256 + 128 + 64 + ... is obviously less than 1024, hence the answer is 2^9.
I'd memorize powers of 2.
I'd memorize powers of 2.
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aspiregmat
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another possible approach...
its a GP series excluding the first 2.
Sn = a(r^n - 1)/r-1
here a = 2
r = 2
n = 8
ans = 2 +2(2^8 - 1) = 2^9
its a GP series excluding the first 2.
Sn = a(r^n - 1)/r-1
here a = 2
r = 2
n = 8
ans = 2 +2(2^8 - 1) = 2^9
