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15 Oct 2009, 17:58
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
70% (02:41) correct
30%
(03:06)
wrong
based on 619
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If the sequence \(x_1\), \(x_2\), \(x_3\), …, \(x_n\), … is such that \(x_1\) = 3 and \(x_{n+1}\) = 2\(x_n\) –1 for n \(\geq\) 1, then \(x_{20}\) – \(x_{19}\)=
A. \(2^{19}\)
B. \(2^{20}\)
C. \(2^{21}\)
D. \(2^{20}\) - 1
E. \(2^{21}\) - 1
A. \(2^{19}\)
B. \(2^{20}\)
C. \(2^{21}\)
D. \(2^{20}\) - 1
E. \(2^{21}\) - 1
04 Aug 2010, 04:25
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If the sequence \(x_1\), \(x_2\), \(x_3\), ..., \(x_n\), ... is such that \(x_1 = 3\) and \(x_{n+1} = 2x_n - 1\) for \(n\geq1\), then \(x_{20} - x_{19}\) equals which of the following?
A. \(2^{19}\)
B. \(2^{20}\)
C. \(2^{21}\)
D. \(2^{20} - 1\)
E. \(2^{21} - 1\)
We have the sequence \(x_1\), \(x_2\), \(x_3\), …, \(x_n,\), …
\(x_1=3\) and \(x_{n+1}=2x_n - 1\) for \(n\geq1\).
If you notice there is a specific pattern in it:
\(x_1=3=2^1+1\)
\(x_2=2x_1-1=5=2^2+1\)
\(x_3=2x_2-1=9=2^3+1\)
...
\(x_n=2^n+1\)
So, \(x_{20}=2^{20}+1\) and \(x_{19}=2^{19}+1\).
\(x_{20}-x_{19}=2^{20}+1-2^{19}-1=2^{20}-2^{19}=2^{19}\)
Answer: A.
Hope it helps.
A. \(2^{19}\)
B. \(2^{20}\)
C. \(2^{21}\)
D. \(2^{20} - 1\)
E. \(2^{21} - 1\)
We have the sequence \(x_1\), \(x_2\), \(x_3\), …, \(x_n,\), …
\(x_1=3\) and \(x_{n+1}=2x_n - 1\) for \(n\geq1\).
If you notice there is a specific pattern in it:
\(x_1=3=2^1+1\)
\(x_2=2x_1-1=5=2^2+1\)
\(x_3=2x_2-1=9=2^3+1\)
...
\(x_n=2^n+1\)
So, \(x_{20}=2^{20}+1\) and \(x_{19}=2^{19}+1\).
\(x_{20}-x_{19}=2^{20}+1-2^{19}-1=2^{20}-2^{19}=2^{19}\)
Answer: A.
Hope it helps.
General Discussion
18 Sep 2010, 11:03
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\(x_{n+1}=2x_n - 1\)
\(x_{n+2}=2*(2x_n - 1) - 1=2^2x_n - (1 + 2)\)
\(x_{n+3}=2*(2^2x_n - (1+2)) - 1=2^3x_n - (1 + 2 + 2^2)\)
\(\vdots\)
\(x_{n+k}=2^kx_n - (1+2+...2^{k-1})=2^k*x_n - (2^k-1)\)
Let n=1
\(x_{k+1} = 2^k*3 - (2^k - 1) = 2^{k+1} + 1\)
So :
\(x_{20} - x_{19} = 2^{20} - 1 - 2^{19} + 1 = 2^{19}\)
\(x_{n+2}=2*(2x_n - 1) - 1=2^2x_n - (1 + 2)\)
\(x_{n+3}=2*(2^2x_n - (1+2)) - 1=2^3x_n - (1 + 2 + 2^2)\)
\(\vdots\)
\(x_{n+k}=2^kx_n - (1+2+...2^{k-1})=2^k*x_n - (2^k-1)\)
Let n=1
\(x_{k+1} = 2^k*3 - (2^k - 1) = 2^{k+1} + 1\)
So :
\(x_{20} - x_{19} = 2^{20} - 1 - 2^{19} + 1 = 2^{19}\)
