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In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X

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Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X [#permalink] New post 27 Aug 2012, 17:30
+1 B

Factorize and you will have solved a great part of the question ;)
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Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X [#permalink] New post 05 Sep 2012, 08:02
I did it like it : as you can see x(1+x(1+x(1+x(1+x)))) X comes 5 times, therefore the max term will be X^5, in the question you see that you want to arrive at X^5 so it means that in the sum of X^n-1...x^n+3 the max term must be X^10 so that it can be x^5(x^5) therefore 10 = 3+n, n=7, timer indicate me 1min 53.

But definitely i had the answer, but i was unable to demonstrate it in that time, it would take more like 5 to 10 minutes.
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Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X [#permalink] New post 19 Sep 2013, 00:09
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Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X [#permalink] New post 19 Sep 2013, 12:34
Could someone please explain to me how that's an infinite sequence? That's what really threw me off.
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Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X [#permalink] New post 19 Sep 2013, 20:48
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mfabros wrote:
Could someone please explain to me how that's an infinite sequence? That's what really threw me off.


The information that it is an infinite sequence doesn't have much to do with the question. You are given this only to tell you that n can take any positive integer value.

An = X^(n-1) + X^n + X^(n+1) + X^(n+2) + X^(n+3) tells you that the nth term is given by plugging in the value of n in this expression. A is not a sequence of 2 or 4 terms but infinite so n can take any value. We found out that the required relation holds when n is 7. We could have just as well got n = 10298 and that would have been fine too since A has infinite terms so any value for n is alright.
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Re: In the infinite sequence A, An = X^(n-1) + X^n + X^(n+1) + X   [#permalink] 19 Sep 2013, 20:48
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