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

Question

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.

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

GMATinsight · Accepted Answer

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

Bunuel · Answer

Yes. Thank you for noticing!

anudeep133 · Answer

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.

Bunuel · Answer

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.

AndyNeedsGMAT · Answer

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

stonecold · Answer

Here product = x^ 1+3+5+...
so we have to indirectly find the sum of => 1+3+5+7+....2k-1
sum = k/2   = k^2 
P.S => i used the formula=> S  = N/2

sahilvijay · Answer

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

Crytiocanalyst · Answer

The power of x is sum of powers of all odd integers 
1+3+5+...+(2k-1) 
which is equal to k^2( the sum of the first k odd integers equals k^2)
Hence IMO D

bumpbot · Answer

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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

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