Question

Derive the formula to find the sum of n-th term of Fibonacci Series.

Answer 1

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..... The “Fibonacci sequence” is defined as a sequence of numbers f_0, f_1, f_2, \cdots such that you have the recursion: f_n = f_{n-1}+f_{n-2}, and the restrictions: f_0 = 1 and f_1 = 1.

Explicitly, the Fibonacci sequence is: 1, 1, 2, 3, 5, 8, 13, 21, … That is, the recursion says that every term is the sum of the previous two.

You can also talk about “generalized Fibonacci sequences”, where these restrictions and/or the recursion are changed. For example: f_n = 2f_{n-1}+3f_{n-2}, with f_0 = 5 and f_1 = 2. This derivation is for the ordinary sequence, but it can be altered to suit any generalized Fibonacci sequence. Here it is:

Every now and again it’s useful to encode a string of numbers in a “generating function“. For obscure (and unimportant to this post) reasons, you can write many functions as infinitely long polynomials. For example: \sin{(x)} = x-\frac{1}{6}x^3+\frac{1}{120}x^5 \cdots. The generating function for a sequence of numbers f_0, f_1, f_2, f_3, \cdots is g(x) = f_0 + f_1 x + f_2 x^2 + f_3 x^3 + \cdots = \sum_{n=0}^\infty f_n x^n. So, sin(x) is the generating function for the sequence 0, 1, 0, -\frac{1}{6},0,\frac{1}{120}, \cdots. If you can find a simple form for this function g, then bully. You’ve got a very straight forward way of writing an infinite string of numbers. The value of x doesn’t have anything to do with anything. The powers of x are really just there to keep the numbers straight. Now check this out!

\begin{array}{ll}g(x) = \sum_{n=0}^\infty f_n x^n\\xg(x) = \sum_{n=0}^\infty f_n x^{n+1} = \sum_{n=1}^\infty f_{n-1} x^n \\x^2 g(x) = \sum_{n=0}^\infty f_n x^{n+2} = \sum_{n=2}^\infty f_{n-2} x^n\end{array}

You can take the recursion and use it to find a relationship between these three slightly different functions. Here’s a good first guess: g(x) = xg(x)+x^2g(x).....

Answer 2

Answer:

Fib(n) = Phin − (−Phi)−n = Phin − ( −phi)n √5 √5

On these pages we use

Phi = √5 + 1 =

1·61803 39887 49894 84820 ... and phi = Phi −1 = 1 = √5 −1 = 0·61803 39887 49894 84820 ... 2 Phi 2

The next version uses just one of the golden section values: Phi, and all the powers are positive:

Fib(n) =

Phin – (–1)n

Phin

sqrt5