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# If the sequence x1, x2, x3, ..., xn, given that x1 = 3 and x

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If the sequence x1, x2, x3, ..., xn, given that x1 = 3 and x  [#permalink]

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Updated on: 09 Jul 2013, 05:52
5
00:00

Difficulty:

45% (medium)

Question Stats:

73% (02:43) correct 27% (02:27) wrong based on 109 sessions

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If the sequence x1, x2, x3, ..., xn, given that x1 = 3 and x(n+1) = 2x(n) - 1 for n >= 1, then x20 - x19 is...

A. 2^(19)
B. 2^(20)
C. 2^(21)
D. 2^(20) - 1
E. 2^(21) - 1

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Originally posted by metallicafan on 18 Sep 2010, 09:41.
Last edited by Bunuel on 09 Jul 2013, 05:52, edited 1 time in total.
Renamed the topic and edited the question.
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18 Sep 2010, 09:46
1
4
metallicafan wrote:
If the sequence X1, X2, X3,...,Xn, given that X1 = 3 and Xn+1 = 2Xn - 1 for n >= 1, then X20 - X19 is...

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 int it:
$$x_1=3=2^1+1$$
$$x_2=5=2^2+1$$
$$x_3=9=2^3+1$$
...
$$x_n=2^n+1$$

So, $$x_{20}=2^{20}+1$$ and $$x_{19}=2^{19}+1$$.

$$x_{20}-x_{19}=2^{20}+1-2^{19}-1=2^{20}-2^{19}=2^{19}$$

Hope it helps.
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18 Sep 2010, 10:03
$$x_{n+1}=2x_n - 1$$
$$x_{n+2}=2*(2x_n - 1) - 1=2^2x_n - (1 + 2)$$
$$x_{n+3}=2*(2^2x_n - (1+2)) - 1=2^3x_n - (1 + 2 + 2^2)$$
$$\vdots$$
$$x_{n+k}=2^kx_n - (1+2+...2^{k-1})=2^k*x_n - (2^k-1)$$

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, given that x1 = 3 and x  [#permalink]

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22 Apr 2018, 20:37
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Re: If the sequence x1, x2, x3, ..., xn, given that x1 = 3 and x &nbs [#permalink] 22 Apr 2018, 20:37
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