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rockrock
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Post subject: The infinite sequence at a1, a2...  Posted: Wed Jun 23, 2010 5:06 pm |
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The infinite sequence a1, a2,...,an is defined such that an = (n+2) / n for all n ≥ 1. What is the product of the first 10 terms of the sequence?
(A) 45
(B) 66
(C) 90
(D) 121
(E) 132
This question was from the sample ONE DAY WORKSHOP question. I'm just curious to know the shortcut for answering this type of sequence problem? I know there is a formula for addition (sum = mean * # of integers), what about for product?
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rockrock
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Post subject: Re: The infinite sequence at a1, a2... Posted: Wed Jun 23, 2010 5:19 pm |
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FYI - the way i solved this was to write out the problem, starting with n=1.
So the first number would be: 3/1....the last number 12/10.
I cancelled out 3 thru 10 in the numerator versus 3-10 in the denominator, leaving me with 12x10 / 2x1 = 66.
That is the correct answer, but is there an easier way? A formula to write out?
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adiagr
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Post subject: Re: The infinite sequence at a1, a2... Posted: Thu Jun 24, 2010 12:37 am |
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rockrock wrote: FYI - the way i solved this was to write out the problem, starting with n=1.
So the first number would be: 3/1....the last number 12/10.
I cancelled out 3 thru 10 in the numerator versus 3-10 in the denominator, leaving me with 12x10 / 2x1 = 66.
That is the correct answer, but is there an easier way? A formula to write out? Hi rock, as you have already seen, series is like: 3/1, 4/2, 5/3, 6/4....... so for ten terms in denominator we have 10!, and in numerator we have 3x4x5x6....x12. Thus numerator is basically (12!/2) so our expression becomes [12!/(10! x 2)]Now 12! = 12 x 11 x 10!. 10! cancels out. [(12x11)/ 2] = 66. Aditya
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rockrock
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Post subject: Re: The infinite sequence at a1, a2... Posted: Thu Jun 24, 2010 12:58 am |
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this is a great explanation. but where in the strategy books would i have found the rule about the sequence that would tell me to divide by 2 in order to get rid of the the last 2 factors in the factorial? I.e. that the numerator is 12!/2 (since we start with 3). I've already passed the number properties and equations strategy guide and that wasn't in either.
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Varshneya
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Post subject: Re: The infinite sequence at a1, a2... Posted: Thu Jun 24, 2010 3:10 am |
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it might be under permutation/combinatrix.
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adiagr
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Post subject: Re: The infinite sequence at a1, a2... Posted: Thu Jun 24, 2010 6:14 am |
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rockrock wrote: this is a great explanation. but where in the strategy books would i have found the rule about the sequence that would tell me to divide by 2 in order to get rid of the the last 2 factors in the factorial? I.e. that the numerator is 12!/2 (since we start with 3). I've already passed the number properties and equations strategy guide and that wasn't in either. Hi Rock, By definition n! = product of all Nos. from 1 to n. numerator is 3x4x5x6x7x8x9x10x11x12. If I multiply the numerator by 2, I will get all Nos. from 1 to 12 and thus I can write this as 12!. And because I am multiplying the numerator by 2, I have to divide by 2, as I cannot change the value of the term That is why expression becomes 12!/2 Hope it is clear now.
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RonPurewal
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Post subject: Re: The infinite sequence at a1, a2... Posted: Thu Jul 08, 2010 8:15 am |
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| ManhattanGMAT Staff |
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the solutions posted above work, but they're a lot of work; you guys shouldn't have to mess around with factorials to solve this thing.
in the vast majority of problems with MULTI-STEP PATTERNS, the easiest and fastest way to solve the problem is to do the following:
1) WRITE OUT a number of steps
2) LOOK for a PATTERN
3) figure out how the pattern will continue
in this problem, if you write out the first few of these fractions, you'll see:
(3/1) x (4/2) x (5/3) x (6/4) x (7/5) x ...
this forum doesn't really allow fractions to be written vertically, but, if you write these fractions out on a piece of paper, the pattern should become clear:
* all of the numerators will cancel out, except the last two;
* all of the denominators will cancel out, except the first two ("1" and "2").
so the only things that don't cancel out are 9 and 10 on top, and 1 and 2 on the bottom.
that leaves (11 x 12)/(1 x 2), or 66. <-- fixed; the original posting contained a mistake
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rockrock
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Post subject: Re: The infinite sequence at a1, a2... Posted: Thu Jul 08, 2010 8:34 am |
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I'm confused. I thought the answer was 66. Which is what I found by canceling out numbers as well....
[editor: that was a mistake. thanks for pointing it out.
i'm not that good with numbers. heh. --ron]
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rohit801
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Post subject: Re: The infinite sequence at a1, a2... Posted: Thu Jul 08, 2010 5:59 pm |
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Going with what Ron said, it makes sense. Just that in this case, the 11 and 12 at the top will be be canceled since the maximun the denominator can get is 10.
But , i think his approach is elegant and simple!!
[editor: nope, i made a mistake. it's now fixed.]
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mschwrtz
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Post subject: Re: The infinite sequence at a1, a2... Posted: Tue Jul 13, 2010 1:10 am |
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| ManhattanGMAT Staff |
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Looks as though everyone appreciated Ron's reasoning, so we're all good?
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