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In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X
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14 Nov 2022, 04:03
14 Nov 2022, 04:03
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Question Stats:
66% (03:51) correct
33% (00:17) wrong based on 3 sessions
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In the infinite sequence A, \(A_n = x^{(n-1)} + x^n + x^{(n+1)} + x^{(n+2)} + x^{(n+3)}\) where x is a positive integer constant. For what value of n is the ratio of \(A_n\) to \(x(1+x(1+x(1+x(1+x))))\) equal to \(x^5\)?
A. 8
B. 7
C. 6
D. 5
E. 4
A. 8
B. 7
C. 6
D. 5
E. 4
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Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X
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14 Nov 2022, 07:45
14 Nov 2022, 07:45
1
Expert Reply
\(An = x^{n-1} + x^n + x^{n+1} + x^{n+2} + x^{n+3}\)
e.g. \(A2 = x + x^2 + x^3 + x^4 + x^5\)
Notice you can only take x common out of all these terms i.e. the smallest term \(x^{n - 1}\)
If \(\frac{An}{{x(1+x(1+x(1+x(1+x))))}} = x^5\), it means the part: (1+x(1+x(1+x(1+x)))) will get canceled from the num and den. Ignore it.
From An, you will be able to take out \(x^6\) common so that \(\frac{x^6}{x}\) gives you \(x^5\)
So smallest term must be \(x^6\) i.e. \(x^{n-1}\). Therefore, n = 7.
Answer: B
e.g. \(A2 = x + x^2 + x^3 + x^4 + x^5\)
Notice you can only take x common out of all these terms i.e. the smallest term \(x^{n - 1}\)
If \(\frac{An}{{x(1+x(1+x(1+x(1+x))))}} = x^5\), it means the part: (1+x(1+x(1+x(1+x)))) will get canceled from the num and den. Ignore it.
From An, you will be able to take out \(x^6\) common so that \(\frac{x^6}{x}\) gives you \(x^5\)
So smallest term must be \(x^6\) i.e. \(x^{n-1}\). Therefore, n = 7.
Answer: B
gmatclubot
Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X [#permalink]
14 Nov 2022, 07:45
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