2006-07-31 23:41:23 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The sum of the first n Fibonacci numbers is:
F1+F2+F3+...+Fn = Fn+2-1
http://www.jimloy.com/algebra/fibo.htm
2006-08-01 00:55:00 · answer #1 · answered by DigitalManic 2 · 0⤊ 0⤋
You may use a formula obtained by solving the difference equation:
f(n) = f(n-1) + f(n-2), f(1)=1 and f(2)=1 . The solution is:
f(n) = ((1+sqrt(5))/2)^n/sqrt(5) - ((1-sqrt(5))/2)^n/sqrt(5)
Ugly?
Yes!
You must use a scientific calculator... or ... software like Maple or Matlab... or Excel formula.
2006-08-01 00:17:19 · answer #2 · answered by vahucel 6 · 0⤊ 0⤋
you would have to solve the recurrence equation t(n-1)+t(n-2) =t(n) its a pretty boring process and u land up with the golden number .
its somehting like (1+5^0.5)/2 n + something....
2006-07-31 23:59:01 · answer #3 · answered by coolelectromagnet 2 · 0⤊ 0⤋
the sequence will be 0, 1, 1, 2, 3, 5, 8, 13,21,34...................
S1=0:S2=1:S3=2:S4=4:S5=7:S6=12:S7=20:S8=33:S9=54:
S10=88
i .lemme try.can't say off hand
2006-08-01 00:04:36 · answer #4 · answered by raj 7 · 0⤊ 0⤋
ΣF(i)=2(F3+F6+F9)=2(2+8+34)=88
is this what you mean??
2006-07-31 23:59:39 · answer #5 · answered by mohamed.kapci 3 · 0⤊ 0⤋