Iterating over an Array Using a “for” Loop.

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This is a very basic question and I would appreciate any help. I've written code to calculate the Fibonacci sequence using a for loop. I want to display the elements of the sequence whose indices are specified in the array “N”. The problem is that all displayed values are the same as the value associated with the first element of “N”.
N=[10 100 1000];
first=1;
second=1;
for i=1:(N-2) %The index has to have two terms removed because it starts with 1 and 1 already
next=first+second; %The current term in the series is a summation of the previous two terms
first=second; %Each term must by iterated upwards by an index of one
second=next; %The term that previously was second is now referred to as next
end
for i=1:length(N)
disp([N(i) next])
end
Thanks
  2 Comments
Donald Hersam
Donald Hersam on 17 Apr 2016
Moved: DGM on 6 Feb 2023
Thanks for the quick response, but what this loop seems to do is tell me the entirety of the series. I was only looking for a response that returns the value at the 10, 100 and 1000th place mark on the series.
Maryam Omar Maabreh
Maryam Omar Maabreh on 5 Feb 2023
Moved: DGM on 6 Feb 2023
for i=start : step : last+1
% ...
end
Where step is the increment to i.
In your case, your looking for
for i=1 : 10 : 10001
i = i*10;
% ...
end

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Accepted Answer

Image Analyst
Image Analyst on 17 Apr 2016
Edited: Image Analyst on 17 Apr 2016
Donald, regarding your new "Answer", you can do it like this:
N=[10 100 1000];
for n = 1 : length(N)
first=1;
second=1;
for k = 3 : N(n) %The index has to have two terms removed because it starts with 1 and 1 already
next = first+second; %The current term in the series is a summation of the previous two terms
first = second; %Each term must by iterated upwards by an index of one
second = next; %The term that previously was second is now referred to as next
end
% Print to command window
next
end

More Answers (1)

MathWorks Support Team
MathWorks Support Team on 2 Sep 2020
Moved: DGM on 6 Feb 2023
Use a “for” loop to calculate the elements of the Fibonacci sequence for the maximum value in “N”. Then, use another “for” loop to display the values by indexing into the calculated sequence.
N = [10 100 1000];
f(1) = 1;
f(2) = 1;
for i = 3:max(N)
f(i) = f(i-1) + f(i-2);
end
for i = 1:numel(N)
disp(f(N(i)))
end

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