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A sequence p1, p2, p3, ... pn is such that p1 = 3 and pn+1 = 2pn − 1

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A sequence p1, p2, p3, ... pn is such that p1 = 3 and pn+1 = 2pn − 1  [#permalink]

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New post 20 Sep 2018, 00:14
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A sequence \(p_1\), \(p_2\), \(p_3\), ... \(p_n\) is such that \(p_1 = 3\) and \(p_{n+1} = 2p_n − 1\) for n ≥ 1, then \(p_{10} − p_9 =\)


(A) \(2^9\)

(B) \(2^{10} − 1\)

(C) \(2^{10}\)

(D) \(2^{11} − 1\)

(E) \(2^{11}\)

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Re: A sequence p1, p2, p3, ... pn is such that p1 = 3 and pn+1 = 2pn − 1  [#permalink]

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New post 20 Sep 2018, 02:14
So p2=2p1-1=2*3-1=5

p3=2p2-1=2*5-1=9

so series will be like

3,5,9,17,33.....

so 5-3=2=2^1
9-5=4=2^2
17-9=8=2^3

so 2nd term -1st term=2^1
3rd term-2nd term=2^2
like wise
10th term-9th term=2^9

so A is answer
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Re: A sequence p1, p2, p3, ... pn is such that p1 = 3 and pn+1 = 2pn − 1  [#permalink]

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New post 20 Sep 2018, 08:29
Bunuel wrote:
A sequence \(p_1\), \(p_2\), \(p_3\), ... \(p_n\) is such that \(p_1 = 3\) and \(p_{n+1} = 2p_n − 1\) for n ≥ 1, then \(p_{10} − p_9 =\)


(A) \(2^9\)

(B) \(2^{10} − 1\)

(C) \(2^{10}\)

(D) \(2^{11} − 1\)

(E) \(2^{11}\)



choices should tell you that the answer has to be in terms of \(2^n\).
keeping this in mind we can work forward..

\(p_{n+1} = 2p_n − 1...........p_{n+1} = p_n+p_n − 1.............p_{n+1} -p_n=p_n − 1\) so \(p_{10} − p_9 =p_9-1\)
so let us find \(p_9\)

now \(p_1=3=2^1+1\)..
\(p_2=2*p_1-1=2*3-1=5=2^2+1\)..
\(p_3=2*p_2-1=2*5-1=9=2^3+1\)
thus we get term for nth term
\(p_n=2^n+1\) and \(p_9=2^9+1\)

and \(p_{10} − p_9 =p_9-1=2^9-1+1=2^9\)

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Re: A sequence p1, p2, p3, ... pn is such that p1 = 3 and pn+1 = 2pn − 1  [#permalink]

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New post 20 Sep 2018, 21:18
Bunuel wrote:
A sequence \(p_1\), \(p_2\), \(p_3\), ... \(p_n\) is such that \(p_1 = 3\) and \(p_{n+1} = 2p_n − 1\) for n ≥ 1, then \(p_{10} − p_9 =\)


(A) \(2^9\)

(B) \(2^{10} − 1\)

(C) \(2^{10}\)

(D) \(2^{11} − 1\)

(E) \(2^{11}\)


To answer such a question one should always write a few terms (as many as you can calculate till 4th or 5th step) and then follow the patterns that question seeks at 10th term

\(p_1 = 3\)
\(p_2 = 2*3-1 = 5\) i.e. \(p_2 − p_1 =5-3 = 2^1\)
\(p_3 = 2*5- 1 = 9\) i.e. \(p_3 − p_2 =9-5 = 2^2\)
\(p_4 = 2*8- 1 = 17\) i.e. \(p_4 − p_3 =17-9 = 2^3\)
\(p_5 = 2*17- 1 = 33\) i.e. \(p_5 − p_4 =33-17 = 2^4\)
.....
.....
..... Following the same pattern we can deduce that \(p_{10} − p_9 = 2^9\)
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Re: A sequence p1, p2, p3, ... pn is such that p1 = 3 and pn+1 = 2pn − 1 &nbs [#permalink] 20 Sep 2018, 21:18
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