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Sequence S is defined as follows: S 1 = 2
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25 Jul 2020, 09:56
25 Jul 2020, 09:56
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Question Stats:
70% (01:14) correct
30% (01:29) wrong based on 20 sessions
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Sequence S is defined as follows:\( S_1 = 2, S_2=2^1, S_3=2^2,......, S_n=2^{n-1}\). What is the sum of the terms is sequence S when \(n=10\) ?
A. \(2^9\)
B. \(2^{10}\)
C. \(2^{16}\)
D. \(2^{35}\)
E. \(2^{37}\)
A. \(2^9\)
B. \(2^{10}\)
C. \(2^{16}\)
D. \(2^{35}\)
E. \(2^{37}\)
Intern
Joined: 25 Jul 2020
Posts: 36
Given Kudos: 0
GRE 1: Q170 V157
Re: Sequence S is defined as follows: S 1 = 2
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Updated on: 27 Jul 2020, 11:03
Updated on: 27 Jul 2020, 11:03
1
The question has a bug. You mention that Sn = 2^(n - 1), but from your description the first term is 2, which is 2^1, not 2^0. And also there isn't quite any correct answer either, because
S_n = a(r^n - 1)/(r - 1). Then S_n = 2(2^n - 1) or (2^n - 1) depending on what a is, and I don't think any of the options fit it.
Edit: missed that the first term is 2, not 1, welp. Answer is \(2^10\) then.
S_n = a(r^n - 1)/(r - 1). Then S_n = 2(2^n - 1) or (2^n - 1) depending on what a is, and I don't think any of the options fit it.
Edit: missed that the first term is 2, not 1, welp. Answer is \(2^10\) then.
Originally posted by Leaderboard on 25 Jul 2020, 10:55.
Last edited by Leaderboard on 27 Jul 2020, 11:03, edited 1 time in total.
Last edited by Leaderboard on 27 Jul 2020, 11:03, edited 1 time in total.
Re: Sequence S is defined as follows: S 1 = 2
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26 Jul 2020, 09:32
26 Jul 2020, 09:32
Leaderboard wrote:
The question has a bug. You mention that Sn = 2^(n - 1), but from your description the first term is 2, which is 2^1, not 2^0. And also there isn't quite any correct answer either, because
S_n = a(r^n - 1)/(r - 1). Then S_n = 2(2^n - 1) or (2^n - 1) depending on what a is, and I don't think any of the options fit it.
S_n = a(r^n - 1)/(r - 1). Then S_n = 2(2^n - 1) or (2^n - 1) depending on what a is, and I don't think any of the options fit it.
When n=10
S = S_1 + S_2 + .... + S_9 + S_10
=> S = 2 + 2^1 + 2^2 + 2^3 + .... + 2^8 + 2^9
=> S - 2 = 2^1 + 2^2 + 2^3 + .... + 2^8 + 2^9
Lets consider RHS as X
X = 2^1 + 2^2 + 2^3 + .... + 2^8 + 2^9
2*X = 2^2 + 2^3 + .... + 2^8 + 2^9 + 2^10
subtract first equation from second
=> 2*X - X = -2^1 + 0 + 0 + 0 + .... + 0 + 0 + 2^10
=> X = 2^10 - 2
S = X + 2 = 2^10
The answer is B.
It would have been better if it was mentioned that S_n = 2^(n-1) for n>1.
