Fibonacci sequence

Let us prove that for every positive integer n this number is a whole number:

The first minute this is a very surprising statement. Gives this formula a whole number with the carried out operations on the irrational numbers in all cases?

In the case n = 0, 1 it shows that the value is 0 and 1, in the case n = 2 we obtain the result after a little counting though, the value is 1.
In the cases n = 3, 4, 5, 6, 7 … the counting is turning into increasingly more difficult, but if someone is not lazy and will not miscalculate it, gets the next results given the following order: 2, 3, 5, 8, …

These numbers may be familiar already though. Yes, there are the first numbers of the Fibonacci sequence. Any number from the third is equal to the sum of the previous two:

We must show that:

Make a common denominator of the fractions:

This is what we wanted.

So the above number F is the n-th number of the Fibonacci sequence given by an explicit form. This is also called Binet formula.