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The value of the product
where the nth factor is
, is(A) 441 (B) 4041 (C) 4410 (D) 4001 (E) 4010
I found this question on one of those Math Competition papers. I don't think you are allowed to use a calculator, or it would probably be easy XD. So yeah... what's the fastest method, just multiplying everything one by one, or is there some scary (and possibly faster) product formula?
Last edited by Toast (2007-01-22 18:29:29)
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The answer is 400 plus 41.
Because when you multiply X by 1 point something, you just add the something fraction of X to X. So for the first two terms, you have 4 and add 5/4 of 4 to 4, which is 4 + 5, or 9.
See how the pattern is just the numbers in the fraction added together.
igloo myrtilles fourmis
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Hmm, that's quite interesting
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John ... that is so well observed!
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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That is a very nice method. I wouldn't have seen to do it that way.
A possibly less complicated method would be to write each bracket as a top-heavy fraction.
Then you get (4/1)*(9/4)*(16/9)*(25/16)*...*(441/400), and you can see that the top term of one fraction cancels with the bottom term of the next, and so you're left with just (441/1) = 441.
Why did the vector cross the road?
It wanted to be normal.
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very very nice !!!
ImPo$$!BLe = NoTH!nG
Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...
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