Question

Help! Fibonacci.

Answer 1

What is the ratio between the 13th and 12th terms in the Fibonacci Sequence?

Round your answer to five decimal places.

Answer 2

\[\approx \phi\]

Answer 3

General term formula is

Answer 4

Well from wikipedia: http://en.wikipedia.org/wiki/Fibonacci_number

We can see that 144 is the 12th term and 233 is the 13th..
Their ration would be 13th/12th term..
We get this: 1.618055556
Rounding this number to five decimal places we get this: 1.61805
which is really close to the golder ration number: http://en.wikipedia.org/wiki/Golden_ratio

Answer 5

Wouldn't it be 1.61806 because we round the 5 to a 6?

Answer 6

Well since the next digit is a "5" you can do whatever you want! Leave it that way or increase the previous digit(making the "5" a "6").

Answer 7

Thanks! Would you mind checking out the question I just posted? @chrisasl

Answer 8

just use the formular above by inputing the 12 and 13 in your caculator

Answer 9

@Ashley10116 Where is it? I can't see it XD

Answer 10

If you have Pascal, you can program it.!! 3 methord in all;

1.
　 　　var 　　b: array[1..40]of longint; 　　i: integer; 　　begin 　　b[1] := 1; 　　b[2] := 1; 　　for i:=3 to 40 do 　　b[i ] := b[i-1] + b[i-2]; 　　for i:=1 to 40 do 　　write(b[i],' '); 　　end. 　

2.
　function fib(n:integer):longint; 　　begin 　　if (n=1) then exit(0); 　　if (n=2) then exit(1); 　　fib:=fib(n-2)+fib(n-1); 　　end; 　　

3.
high accuracy
　　program fzu1060; 　　type arr=array[0..1001]of integer; 　　var a,b,c:arr; 　　i,j,k,n:integer; 　　procedure add(var a,b,c:arr); 　　begin 　　fillchar(c,sizeof(c),0); 　　c[0]:=b[0]; 　　for i:=1 to c[0] do 　　c[i]:=a[i]+b[i]; 　　for i:=1 to c[0] do 　　begin 　　c[i+1]:=c[i+1]+(c[i] div 10); 　　c[i]:=c[i] mod 10; 　　end; 　　if c[c[0]+1]>0 then 　　begin 　　inc(c[0]); 　　inc(c[c[0]+1],c[c[0]] div 10); 　　c[c[0]]:=c[c[0]] mod 10; 　　end; 　　a:=b; b:=c; 　　end; 　　begin 　　assign(input,'d:\input.txt'); 　　assign(output,'d:\output.txt'); 　　reset(input); 　　rewrite(output); 　　while not eof do 　　begin 　　readln(n); 　　fillchar(a,sizeof(a),0); 　　fillchar(b,sizeof(b),0); 　　a[0]:=1; a[1]:=1; 　　b[0]:=1; b[1]:=1; 　　if n=1 then write(1) 　　else 　　if n=2 then write(1) 　　else 　　begin 　　for k:=3 to n do 　　add(a,b,c); 　　for k:=c[0] downto 1 do 　　write(c[k]); 　　end; 　　writeln; 　　end; 　　close(input); 　　close(output); 　　end.

Answer 11

@FoolForMath @satellite73 am i right?