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c23
Jan 7, 2010, 01:22 PM
1,1,2,3,5,8-- give the nth term
And
2,9,16,23,30,37,44- give the nth term..
Help me tnxxxxxxxxx:confused:

galactus
Jan 7, 2010, 01:43 PM
1,1,2,3,5,8-- give the nth term

This is the Fibonacci sequence. Each term is the sum of the previous two terms. To find the nth Fibonacci number, use what is known as Binet's formula:

F(n)=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5 }}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right)

Plug in any n value and you will get that Fibonacci number. Let's say we want the 15th Fibonacci number. Plug in n=15 into the above formulla and get 610

1,1,2,3,5,8,13,21,33,55,89,144,233,377,610

But, I do not believe we must go that far.

To wit: F(n)=F_{n-2}+F_{n-1}

See? The sum of the two previous numbers.



2,9,16,23,30,37,44- give the nth term..

These numbers just have a difference of 7. Surely you saw that?

galactus
Jan 8, 2010, 12:04 PM
Take note, the Binet formula is based on the Golden Ratio. See the \frac{\sqrt{5}-1}{2}, \;\ \frac{\sqrt{5}+1}{2}? That's the Golden Ratio.

Remember, the \lim_{n\to \infty}\frac{F_{n+1}}{F_{n}}=\text{The Golden Ratio}

What this means is if we divide a Fibonacci number by the previous one, the higher we go the closer it gets to the Golden Ratio.

i.e \frac{13}{8}=1.625

\frac{21}{13}=1.6153846...

\frac{34}{21}=1.6190476....

Jumping ahead: \frac{377}{233}=1.61802575....

and on and on. See how it converges to the Golden Ratio?

The Fibonacci numbers have so many properties and are very interesting.

InfoJunkie4Life
Jan 8, 2010, 02:16 PM
The seven one can be simply

7n + 2

To find the nth sequence

c23
Jan 16, 2010, 04:28 PM
Tnx alll :)