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Dec 16th, 2003, 04:37 PM
#1
Thread Starter
Frenzied Member
Just for fun - number sequence
The 1st term in a geometric(?) number sequence is 2. The 5th term is 14.
what is the formula for the sequence?
edit: hmm.. maybe you need one more term. OK, the 57th is 170.
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Dec 16th, 2003, 06:33 PM
#2
Perhaps you could refresh my memory and illustrate what the form of a geometric series looks like?
Is it something like:
Xi = A*X3(i-1) + B*X2(i-1) + C*X(i-1) + D
???
-Lou
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Dec 16th, 2003, 06:39 PM
#3
Thread Starter
Frenzied Member
I'm not sure this is a geometric sequency, that's why I put it in brackets.
I think a geometric sequence is one where you multiply the number by a number which is constant. Then it can add or subtract number afterwards.
eg
Un = 2n+1
would give
3, 5, 7, 9 ...
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Dec 17th, 2003, 06:15 AM
#4
transcendental analytic
thats the aritmetic series
the geometric series goes exponential:
a+aq+aq^2+..+aq^n = a(1-q^n)/(1-q), q/=1
a=2
2q^5=15 gives q=7.5^0.2~1.496
you don't really need 57'th term, but lets see if you made this up..
2*7.5^(0.2*57)~18911605435 obviously not 170
Use  
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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Dec 17th, 2003, 06:23 AM
#5
transcendental analytic
lets give the aritmetic one a shot (sorry i didn't notice it was you who posted that)
the n'th number should be a+nd where a is the first
2+5d=14 <=> d= 12/5 = 2.4
2+57d=170 <=> d=168/57 = 2.9
doesn't work out...
Use  
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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Dec 17th, 2003, 09:27 AM
#6
Xn = 3n - 1
So,
N=1 => X=2
N=5 => X=14
N=57 => X=170 (3*57 - 1 =171 - 1 = 170)
Its just a linear equation.
-Lou
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Dec 17th, 2003, 05:17 PM
#7
Thread Starter
Frenzied Member
Your right NotLKH, how did you get it?
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Dec 17th, 2003, 05:29 PM
#8
Addicted Member
14 - 2 = 12 (Difference between 5th term and 1st term)
12 / 4 = 3 (Divide by 4 as there is a difference of 4 terms between the first and fifth)
=> coefficient of n is 3
So far Un = 3n
Using that for the first term gives 3.
2 - 3 = - 1
=> constant is - 1
Therefore:
Un = 3n - 1
(This method only works for linear equations though)
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Dec 18th, 2003, 08:10 PM
#9
And on the TI-83 you could've used a linear regression, although Celest's way is better
The time you enjoy wasting is not wasted time.
Bertrand Russell
<- Remember to rate posts you find helpful.
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Dec 20th, 2003, 04:55 AM
#10
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Dec 20th, 2003, 03:41 PM
#11
transcendental analytic
the aritmetic series first element would be -1, not 2..
Use  
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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Dec 21st, 2003, 11:59 PM
#12
Hyperactive Member
Tn = 3n - 1
The first term, T1 is 2, the 0th term, T0, would be -1.
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Dec 22nd, 2003, 04:47 AM
#13
transcendental analytic
there is no 0th term, in math or in english. You start counting from 1
Use  
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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