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

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Joined: 02 Sep 2009
Posts: 50076
A sequence p1, p2, p3, ... pn is such that p1 = 3 and pn+1 = 2pn − 1  [#permalink]

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20 Sep 2018, 00:14
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45% (medium)

Question Stats:

68% (02:10) correct 32% (02:12) wrong based on 41 sessions

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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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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

Math Expert
Joined: 02 Aug 2009
Posts: 6985
Re: A sequence p1, p2, p3, ... pn is such that p1 = 3 and pn+1 = 2pn − 1  [#permalink]

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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 $$p_{10} − p_9 =p_9-1=2^9-1+1=2^9$$

A
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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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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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