LET THE SERIES BE = 0,1,1,2,3,5,8,......,n,m
AND WE NEED TO FIND THE SUM OF FIBO. NOS. TILL m
THEN LETS FIRST CONSIDER
1,1,2,2+1,2(2)+1,3(2)+2(1)
WE OBSERVE THAT F(n+2) = 2{F(n)} + 1{F(n-1)}
THE SUM OF FIBO NOS. TILL m CAN BE GIVEN = F(n+2)-1
BY THE COMBINING THE FORMULAS WE GET:
S(0,1,1,2,3,5,8,......,n,m) = 2(m) + n -1
THERFORE
S(0,1,1,2,3,5,8,13,21,34) = 2(34) + 21 - 1
= 68 + 21 - 1
= 89-1
= 88
2006-08-03 06:19:24
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answer #1
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answered by Yo! Mathematics 2
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S(1) = Sum ( 1 ) = 1
S(2) = Sum ( 1, 1 ) = 2
S(3) = Sum ( 1, 1, 2 ) = 4
S(4) = Sum ( 1, 1, 2, 3 ) = 7
S(5) = Sum ( 1, 1, 2, 3, 5 ) = 12
S(6) = Sum ( 1, 1, 2, 3, 5, 8 ) = 20
S(7) = Sum ( 1, 1, 2, 3, 5, 8, 13 ) = 33
S(8) = Sum ( 1, 1, 2, 3, 5, 8, 13, 21 ) = 54
S(9) = Sum ( 1, 1, 2, 3, 5, 8, 13, 21, 34 ) = 88
S(10) = Sum ( 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 ) = 143
etc.
Notice how S(n) = F(n+2) - 1
So:
S(10) = F(12) - 1
That would be the 12th fibonacci number minus 1.
F(12) = 144
So:
S(10) = 143
You can also use Binet's formula (closed form) to calculate F(n).
Take the golden ratio: Φ = (1 + √5)/2 ≈ 1.618...
F(n) = round ( Φⁿ / √5 )
F(12) = round ( 1.618^12 / sqrt(5) )
F(12) = 144
So:
S(n) = round [ Φ^(n+2) / √5 ] - 1
S(10) = round [ 1.618^12) / √5 ] - 1
S(10) = 143
2006-08-03 05:03:30
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answer #2
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answered by Puzzling 7
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Since 1+10=11, 2+9=11... & you started with an odd number & ended w/ an even (1-10), do (half of ten) times 11, aka 5x11=55. Any other math questions that are Alg. 1 or beginning geometry & down, feel free to email me @ my nickname! :)
2006-08-03 04:55:32
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answer #3
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answered by beckaroo_messer 2
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just arrange the numbers in order Ex: 10 14 24 38
2006-08-03 04:47:48
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answer #4
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answered by Valkyrie 4
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for one thing to get a sum is to add so i looked up this...
http://encyclopedia.thefreedictionary.com/Fibonacci+
hope it helps. LOL
2006-08-03 04:52:12
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answer #5
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answered by '~*Elizabeth*~' 1
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Is it that difficult to add just 10 numbers?
2006-08-03 04:48:48
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answer #6
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answered by Joy M 7
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you're able to use a formulation obtained by using fixing the version equation: f(n) = f(n-one million) + f(n-2), f(one million)=one million and f(2)=one million . the answer is: f(n) = ((one million+sqrt(5))/2)^n/sqrt(5) - ((one million-sqrt(5))/2)^n/sqrt(5) grotesque? sure! you ought to apply a scientific calculator... or ... utility like Maple or Matlab... or Excel formulation.
2016-10-01 10:32:52
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answer #7
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answered by ? 3
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Yeah, click here to get the answer
http://en.wikipedia.org/wiki/Fibonacci_sequence
2006-08-03 04:50:17
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answer #8
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answered by Anonymous
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I sure can't!
2006-08-03 04:47:32
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answer #9
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answered by Love always, Kortnei 6
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