writing long equation with correct answer

Hello everyone, this is not hw ;) I'm self-learner. I'ev faced this equation but couldn't get the final answer correctly. the final answer is 55, but I have different answers. I will attach the equation and my trails. Apology for my low level in english :'(

my trails:

n=10;
fn= [(1/sqrt(5))].* [(1+sqrt(5)/(2)).^n - (1-sqrt(5)/(2).^n)]

fn = 812.0928

and this one

fn= 1/sqrt(5).* (1+sqrt(5)/(2).^n - (1-sqrt(5)/(2).^n))

fn = 0.0020

Answer 1

[Chunru]

Note that in matlab "( )" are very different from "[ ]":

n=10;
fn= 1/sqrt(5) * ( ( (1+sqrt(5)) / 2 )^n - (1-sqrt(5)/2).^n )

fn = 55.0036

[Walter Roberson]

However, [] is compatible enough with () that it was not the source of the problem.

[Chunru]

Wrongly intended use of [] make the code error prone.

[Abdullah Sultan]

thank you so much , I still watching videos and reading books, I hope to impove

[Abdullah Sultan]

just to understand matlab proccess, why in the first (1+sqrt(5)) part we should put () more than the second part (1-sqrt(5)/2) even though they are the same side

to be more clear I make the () bold to mention why here we should make more () the other side

fn= 1/sqrt(5) * ( ( (1+sqrt(5)) / 2 )^n - (1-sqrt(5)/2).^n )

Answer 2

[Walter Roberson]

fn= [(1/sqrt(5))].* [(1+sqrt(5)/(2)).^n - (1-sqrt(5)/(2).^n)]

Notice that in the first sub-expression it is the 1+ expression that is being raised to n, but in the second sub-expression, it is only the 2 that is being raised to the n.

[Abdullah Sultan]

thank you for drawing my attention to this point.