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Math Expert V
Joined: 02 Sep 2009
Posts: 58445
For any given x and any positive integer n, the nth term in sequence S  [#permalink]

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10 00:00

Difficulty:   65% (hard)

Question Stats: 64% (02:25) correct 36% (02:20) wrong based on 219 sessions

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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

Kudos for a correct solution.

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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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4
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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

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Math Expert V
Joined: 02 Sep 2009
Posts: 58445
Re: For any given x and any positive integer n, the nth term in sequence S  [#permalink]

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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

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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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.
Math Expert V
Joined: 02 Sep 2009
Posts: 58445
Re: For any given x and any positive integer n, the nth term in sequence S  [#permalink]

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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.

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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GMAT 1: 610 Q42 V32 Re: For any given x and any positive integer n, the nth term in sequence S  [#permalink]

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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.

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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GRE 1: Q169 V154 Re: For any given x and any positive integer n, the nth term in sequence S  [#permalink]

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Here product = x^ 1+3+5+...
so we have to indirectly find the sum of => 1+3+5+7+....2k-1
sum = k/2 [1+2k-1] = k^2
P.S => i used the formula=> S[n] = N/2 [first term + last term]
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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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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  [#permalink]

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_________________ Re: For any given x and any positive integer n, the nth term in sequence S   [#permalink] 26 Aug 2018, 08:59
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