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
[spoiler]OA=E[/spoiler]
Source: Manhattan GMAT
For any given x and any positive integer n
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For any sequence question, it is usually a good idea to work out the first few terms until you understand the structure of the sequence. Here, if
s_n = x^(2n - 1)
then plugging in n =1, 2 and 3 we find:
s_1 = x^1
s_2 = x^3
s_3 = x^5
and so on. We want to multiply these terms together, which means we'll add the exponents. So the question is just asking: what is the sum of the odd numbers from 1 up to 2k - 1?
There are a couple of ways to finish the problem. We just worked out that, when k=3, the first three terms are x^1, x^3, and x^5, which have a product of x^9. So when k=3, the answer to the question is 9. Plugging k=3 into each answer choice, only E works, so E must be right (you need to try k=3 here, because if you try k=1 or k=2, you still have at least two candidate answer choices). Or we can do this algebraically. We want to add:
1, 3, 5, ..., 2k-1
This is an equally spaced list, increasing by 2 each time, so the mean of the list is equal to the average of the smallest and largest terms. So the mean of the list is (1 + 2k - 1)/2 = k. To find the number of terms n in any equally spaced list, we can use the formula:
n = (range / space) + 1
and here the range is (2k-1)-1 = 2k-2, and the spacing is 2, so
n = (2k-2)/2 + 1 = k-1+1 = k
So the mean of the list is k, the number of terms is k, and from the definition of the average, sum = avg*n, so sum = k*k = k^2.
s_n = x^(2n - 1)
then plugging in n =1, 2 and 3 we find:
s_1 = x^1
s_2 = x^3
s_3 = x^5
and so on. We want to multiply these terms together, which means we'll add the exponents. So the question is just asking: what is the sum of the odd numbers from 1 up to 2k - 1?
There are a couple of ways to finish the problem. We just worked out that, when k=3, the first three terms are x^1, x^3, and x^5, which have a product of x^9. So when k=3, the answer to the question is 9. Plugging k=3 into each answer choice, only E works, so E must be right (you need to try k=3 here, because if you try k=1 or k=2, you still have at least two candidate answer choices). Or we can do this algebraically. We want to add:
1, 3, 5, ..., 2k-1
This is an equally spaced list, increasing by 2 each time, so the mean of the list is equal to the average of the smallest and largest terms. So the mean of the list is (1 + 2k - 1)/2 = k. To find the number of terms n in any equally spaced list, we can use the formula:
n = (range / space) + 1
and here the range is (2k-1)-1 = 2k-2, and the spacing is 2, so
n = (2k-2)/2 + 1 = k-1+1 = k
So the mean of the list is k, the number of terms is k, and from the definition of the average, sum = avg*n, so sum = k*k = k^2.
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We see that each term has base x, and we know that when we multiply terms with the same base, we add the exponents. We see that S_1(x) has 1, S_2(x) has 3, ..., and S_k(x) has 2k - 1 as exponents. So the sum of these exponents is:VJesus12 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
[spoiler]OA=E[/spoiler]
Source: Manhattan GMAT
1 + 3 + ... + 2k - 1
We see that this is the sum of the odd integers from 1 to (2k - 1), inclusive. Since the terms in this sum are evenly spaced, we can use the formula Sum = Average x Quantity. The Average = (1 + 2k - 1)/2 = k, and Quantity = k; therefore, the Sum = k x k = k^2.
Answer: E
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