KweLi
Full Member
Posts: 29
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Post by KweLi on Mar 27, 2002 1:15:51 GMT -5
Does anyone know how to solve 36 and 40? for 36 i tried ln-ing the function but couldnt figure out anything else. i know it looks like (1+x/n)^n but the T is screwing me up. As for 40, i know how to show that c is a lower bound and 2d is an upper bound and to showthe function is increasing, however, I dont know how to show that it goes to d.
Oh one more thing, for 10.3 # 58, i cant figure out what a(n) equation is. i know its something over 3^n-1 but i cant see a relationship with the numerators.
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JebusMeany
Junior Member
Who is Jebus Meany?
Posts: 22
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Post by JebusMeany on Mar 27, 2002 19:10:48 GMT -5
numerator is (n-1)^2 + 3^(n-1)
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Post by LaevisX on Mar 27, 2002 19:57:28 GMT -5
numerator is (n-1)^2 + 3^(n-1) Unless I'm mistaken, a(4)=40/27 and a(5)=121/81, which doesn't work with the numerator you gave -- it works up to a(3). I believe the numerator is a function of 3 n.
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KweLi
Full Member
Posts: 29
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Post by KweLi on Mar 27, 2002 20:31:52 GMT -5
Yeah ive been trying all day and i cant deem to figure out the pattern. ive already tried that and like LaevisX says the 4th and fifth terms dont work for it....
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Yin
Junior Member
umm...let me meditate and get my answer...umm
Posts: 21
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Post by Yin on Mar 27, 2002 23:20:10 GMT -5
do we really need to find the pattern for 58? or are there formulas that lets us use the pinch thrm? (this is a question...not a hint....) bee ntrying all sorts of numerators......but all only fit at most 3 terms...
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Post by Rogue_Knight on Mar 27, 2002 23:36:57 GMT -5
Are you allowed to use sigma notation? because if you can, the general case for 58 would be: (3^0+3^1+...+3^n)/3^(n-1)
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Post by LaevisX on Mar 27, 2002 23:41:09 GMT -5
do we really need to find the pattern for 58? or are there formulas that lets us use the pinch thrm? (this is a question...not a hint....) bee ntrying all sorts of numerators......but all only fit at most 3 terms... Ok, at the risk of giving the whole thing away, here's a big tip: a(n) can be written as (x/y)+1. Notice that in all cases x=(y-1)/2, and the original post tells you what y should be. Figure it out from there ;D
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Post by Rogue_Knight on Mar 27, 2002 23:59:17 GMT -5
Hmmm... it turns into a geometric sum. How wonderful
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KweLi
Full Member
Posts: 29
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Post by KweLi on Mar 28, 2002 0:56:50 GMT -5
Thanks for that... i ve been staring at this question for too long trying to figure that out.
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