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# The sequence Xn is defined as follows: Xn = 2X(n-1) - 1

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Senior Manager
Status: Gathering chakra
Joined: 05 Feb 2018
Posts: 369
The sequence Xn is defined as follows: Xn = 2X(n-1) - 1  [#permalink]

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19 Jun 2019, 15:56
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55% (hard)

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61% (02:18) correct 39% (02:13) wrong based on 18 sessions

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The sequence $$X_{n}$$ is defined as follows: $$X_{n} = 2X_{(n-1)}-1$$ whenever n is an integer greater than 1. If $$X_1=3$$, what is the value of $$X_{20} - X_{19}$$?

A) $$2^{16}$$
B) $$2^{17}$$
C) $$2^{18}$$
D) $$2^{19}$$
E) $$2^{20}$$
Director
Joined: 19 Oct 2018
Posts: 709
Location: India
Re: The sequence Xn is defined as follows: Xn = 2X(n-1) - 1  [#permalink]

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19 Jun 2019, 17:36
1
$$X_1$$=$$2^1+1$$
$$X_2$$= $$2* (2+1)-1$$= $$2^2+1$$
and so on
We can generalize any term of the sequence as
$$X_n$$=$$2^n+1$$
$$X_{20}$$=$$2^{20}+1$$
$$X_{19}$$=$$2^{19}+1$$

$$X_{20}$$-$$X_{19}$$=$$[2^{20}+1]$$- $$[2^{19}+1]$$
$$X_{20}$$-$$X_{19}$$= $$2^{20}-2^{19}$$=$$2^{19}$$

energetics wrote:
The sequence $$X_{n}$$ is defined as follows: $$X_{n} = 2X_{(n-1)}-1$$ whenever n is an integer greater than 1. If $$X_1=3$$, what is the value of $$X_{20} - X_{19}$$?

A) $$2^{16}$$
B) $$2^{17}$$
C) $$2^{18}$$
D) $$2^{19}$$
E) $$2^{20}$$
Senior Manager
Joined: 16 Jan 2019
Posts: 312
Location: India
Concentration: General Management
WE: Sales (Other)
The sequence Xn is defined as follows: Xn = 2X(n-1) - 1  [#permalink]

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19 Jun 2019, 17:42
energetics wrote:
The sequence $$X_{n}$$ is defined as follows: $$X_{n} = 2X_{(n-1)}-1$$ whenever n is an integer greater than 1. If $$X_1=3$$, what is the value of $$X_{20} - X_{19}$$?

A) $$2^{16}$$
B) $$2^{17}$$
C) $$2^{18}$$
D) $$2^{19}$$
E) $$2^{20}$$

$$X_{1}=3$$, $$X_{2}=5$$, $$X_{3}=9$$, $$X_{4}=17$$, $$X_{5}=33$$

We can see that the $$n_{th}$$ term of the sequence $$X_{n}$$ $$=2^n + 1$$

Therefore $$X_{20}$$$$=2^{20}+1$$ and $$X_{19}$$$$=2^{19}+1$$

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

$$=2^{19}(2-1)=2^{19}$$

Hit Kudos if this helped!

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The sequence Xn is defined as follows: Xn = 2X(n-1) - 1   [#permalink] 19 Jun 2019, 17:42
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