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If the sequence x_1, x_2, x_3, …, x_n, ... is such that x_1

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If the sequence x_1, x_2, x_3, …, x_n, ... is such that x_1 [#permalink] New post   15 Oct 2009, 17:58
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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
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If the sequence x1, x2, x3, ... xn, ... is such that x1 = 3 and xn+1 = [#permalink] New post   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.
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Re: If the sequence x1, x2, x3, ... xn, ... is such that x1 = 3 and xn+1 = [#permalink] New post   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 :

\(x_{20} - x_{19} = 2^{20} - 1 - 2^{19} + 1 = 2^{19}\)
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Re: If the sequence x1, x2, x3, ... xn, ... is such that x1 = 3 and xn+1 = [#permalink] New post   14 Jul 2015, 18:29
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x1 = 3
x2 = 2(3)-1 = 5
x3 = 2(5)-1 = 9
x4 = 2(9)-1 = 17

we can notice the general pattern in this series:
(x2-x1) = 2 = 2^1
(x3-x2)= 4 = 2^2
(x4-x3) = 8 = 2^3... and so on.

So, (x^20 - x^19) = 2^(19). Ans (A).
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Re: If the sequence x_1, x_2, x_3, …, x_n, ... is such that x_1 [#permalink] New post   20 Nov 2015, 02:30
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eresh wrote:
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


\(x_1\) = 3
\(x_2\) = 2\(x_1\) –1 = 5
\(x_3\) = 2\(x_2\) –1 = 9
\(x_4\) = 2\(x_3\) –1 = 17
\(x_5\) = 2\(x_4\) –1 = 33

On trying to find a series, this translates to
\(x_1\) = \(2^1 + 1\)
\(x_2\) = \(2^2 + 1\)
\(x_3\) = \(2^3 + 1\)
\(x_4\) = \(2^4 + 1\)
\(x_5\) = \(2^5 + 1\)

Hence \(x_{20}\) = \(2^{20} + 1\) and \(x_{19}\) =\(2^{19} + 1\)

\(x_{20}\) - \(x_{19}\) = (\(2^{20}\) + 1) - (\(2^{19}\) + 1) = \(2^{20}\) - \(2^{19}\) =\(2^{19}\) (2 - 1) = \(2^{19}\)
Option A
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Re: If the sequence x_1, x_2, x_3, , x_n, ... is such that x_1 [#permalink] New post   29 Sep 2023, 08:37
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Re: If the sequence x_1, x_2, x_3, , x_n, ... is such that x_1 [#permalink]
29 Sep 2023, 08:37
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