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2^1 + 2^2 + 2^3 + 2^4 + 2^5
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05 Nov 2017, 01:41
05 Nov 2017, 01:41
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Question Stats:
99% (01:14) correct
0% (00:00) wrong based on 7 sessions
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\(2^1 + 2^2 + 2^3 + 2^4 + 2^5=\)
(A) 2^14
(B) 2^15
(C) 2^9 + 2^5
(D) 2^7 – 2
(E) 2^6 – 2
Kudos for correct solution.
(A) 2^14
(B) 2^15
(C) 2^9 + 2^5
(D) 2^7 – 2
(E) 2^6 – 2
Kudos for correct solution.
Re: 2^1 + 2^2 + 2^3 + 2^4 + 2^5
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05 Nov 2017, 06:32
05 Nov 2017, 06:32
1
I think there are many strategies to face this kind of questions, mine is the following: factorizing 2 repetitively from the sum of powers of 2, i.e.
[m]2+2^2+2^3+2^4+2^5 = 2(1+2+2^2+2^3+2^4) = 2(1+2(1+2+2^2+2^3)) = \dots = 2(1+2(1+2(1+2(1+2)))) = 62/[m].
Then, knowing the first powers of 2 it is easy to see that 2^6 = 64, so that 62 = 2^6-2.
Answer E
[m]2+2^2+2^3+2^4+2^5 = 2(1+2+2^2+2^3+2^4) = 2(1+2(1+2+2^2+2^3)) = \dots = 2(1+2(1+2(1+2(1+2)))) = 62/[m].
Then, knowing the first powers of 2 it is easy to see that 2^6 = 64, so that 62 = 2^6-2.
Answer E
Re: 2^1 + 2^2 + 2^3 + 2^4 + 2^5
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21 Aug 2021, 11:22
21 Aug 2021, 11:22
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