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For any given x and any positive integer n, the nth term in sequence S

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For any given x and any positive integer n, the nth term in sequence S  [#permalink]

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New post 14 Sep 2015, 05:06
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For any given x and any positive integer n, the nth term in sequence S is defined by the equation \(S_n(x) = x^{(2n-1)}\). The product of all terms in sequence S from \(S_1(x)\) through \(S_k(x)\) for any positive integer k is equal to x raised to what power?

(A) k
(B) 2k – 1
(C) 2k
(D) (k^2+ 1)/2
(E) k^2


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Re: For any given x and any positive integer n, the nth term in sequence S  [#permalink]

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New post 14 Sep 2015, 06:18
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Bunuel wrote:
For any given x and any positive integer n, the nth term in sequence S is defined by the equation \(S_n(x) = x^(2n-1)\). The product of all terms in sequence S from \(S_1(x)\) through \(S_k(x)\) for any positive integer k is equal to x raised to what power?

(A) k
(B) 2k – 1
(C) 2k
(D) (k^2+ 1)/2
(E) k^2


Kudos for a correct solution.


Bunuel: by \(S_n(x) = x^(2n-1)\) I hope you meant \(S_n(x) = x^{(2n-1)}\)

@x=1, \(S_1(x) = x^{(2*1-1)} = x\)
@x=2, \(S_2(x) = x^{(2*2-1)} = x^3\)
@x=3, \(S_3(x) = x^{(2*3-1)} = x^5\)

for k = 2
\(S_1(x) * S_2(x) = x*x^3 = x^4\) i.e. Power of x = 4


for k = 3
\(S_1(x) * S_2(x) * S_3(x) = x*x^3*x^5 = x^9\) i.e. Power of x = 9


Looking at two results we conclude that Power of x is always square of k

Answer: option E
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Re: For any given x and any positive integer n, the nth term in sequence S  [#permalink]

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New post 14 Sep 2015, 06:20
GMATinsight wrote:
Bunuel wrote:
For any given x and any positive integer n, the nth term in sequence S is defined by the equation \(S_n(x) = x^(2n-1)\). The product of all terms in sequence S from \(S_1(x)\) through \(S_k(x)\) for any positive integer k is equal to x raised to what power?

(A) k
(B) 2k – 1
(C) 2k
(D) (k^2+ 1)/2
(E) k^2


Kudos for a correct solution.


Bunuel: by \(S_n(x) = x^(2n-1)\) I hope you meant \(S_n(x) = x^{(2n-1)}\)

@x=1, \(S_1(x) = x^{(2*1-1)} = x\)
@x=2, \(S_2(x) = x^{(2*2-1)} = x^3\)
@x=3, \(S_3(x) = x^{(2*3-1)} = x^5\)

for k = 2
\(S_1(x) * S_2(x) = x*x^3 = x^4\) i.e. Power of x = 4


for k = 3
\(S_1(x) * S_2(x) * S_3(x) = x*x^3*x^5 = x^9\) i.e. Power of x = 9


Looking at two results we conclude that Power of x is always square of k

Answer: option E


Yes. Thank you for noticing!
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Re: For any given x and any positive integer n, the nth term in sequence S  [#permalink]

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New post 14 Sep 2015, 06:25
Bunuel wrote:
For any given x and any positive integer n, the nth term in sequence S is defined by the equation \(S_n(x) = x^{(2n-1)}\). The product of all terms in sequence S from \(S_1(x)\) through \(S_k(x)\) for any positive integer k is equal to x raised to what power?

(A) k
(B) 2k – 1
(C) 2k
(D) (k^2+ 1)/2
(E) k^2


Kudos for a correct solution.


Solution: The power of x is sum of powers of \(S_1(x)\) through \(S_k(x)\).
1+3+5+...+(2k-1) which is equal to k^2.

Option E.
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Re: For any given x and any positive integer n, the nth term in sequence S  [#permalink]

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New post 20 Sep 2015, 21:29
Bunuel wrote:
For any given x and any positive integer n, the nth term in sequence S is defined by the equation \(S_n(x) = x^{(2n-1)}\). The product of all terms in sequence S from \(S_1(x)\) through \(S_k(x)\) for any positive integer k is equal to x raised to what power?

(A) k
(B) 2k – 1
(C) 2k
(D) (k^2+ 1)/2
(E) k^2


Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

Start by listing simple terms of sequence S in order.

\(S_1(x) = x^{(2(1)-1)} = x^1 = x\)
\(S_2(x) = x^{(2(2)-1)} = x^3\)
\(S_3(x) = x^{(2(3)-1)} = x^5\)

By this point, we would hopefully notice that the exponents are the positive odd numbers in order.

Next, we need to consider the product of several of these terms, starting from \(S_1(x)\) on up to the kth term.

If k = 1, then the product of all the terms is simply x (since there is only the first term). The exponent here is 1. If we check the answer choices at this point, we could eliminate (C), because if we plug in k = 1, we get an incorrect result here (that is, 2). All the other answer choices give us 1, the correct result, so we have to keep going.

If k = 2, then the product of the first two terms is \(x*x^3 = x^4\). So we need to get 4 out if we plug in k = 2. Of the remaining answer choices, only (E) works. We can stop here.

If we really wanted to, we could check another value of k, or we could notice that we’re summing up the first k odd integers (because we’re multiplying the terms, which have the same base x and those integers as exponents). The sum of the first k odd integers equals k^2.

The correct answer is E.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: For any given x and any positive integer n, the nth term in sequence S  [#permalink]

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New post 04 Mar 2016, 09:11
Bunuel wrote:
For any given x and any positive integer n, the nth term in sequence S is defined by the equation \(S_n(x) = x^{(2n-1)}\). The product of all terms in sequence S from \(S_1(x)\) through \(S_k(x)\) for any positive integer k is equal to x raised to what power?

(A) k
(B) 2k – 1
(C) 2k
(D) (k^2+ 1)/2
(E) k^2


Kudos for a correct solution.


The series looks like : x^1 + x^3 + x^5 ..... x^(2k-1) = x ^ {1+3+5+..... + (2k-1)}
We know sum of n consecutive odd numbers = n^2 => (2k-1)^2
Clearly, x is raised to k^2
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Re: For any given x and any positive integer n, the nth term in sequence S  [#permalink]

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New post 09 Mar 2016, 04:17
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Re: For any given x and any positive integer n, the nth term in sequence S  [#permalink]

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New post 13 Aug 2017, 12:06
Ans is E

let us consider 3 terms
s1= x
s2=x^3
s3=x^5
s1 x s2 x s3 = x^1 . x^3 . x^5 = x^9

put value of k = 3 in answer only E gives 9 since 3^2 is 9
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Re: For any given x and any positive integer n, the nth term in sequence S &nbs [#permalink] 13 Aug 2017, 12:06
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