Effacer les filtres
Effacer les filtres

Help me with this question please....

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Mohammed Safwat
Mohammed Safwat le 11 Déc 2017
Modifié(e) : Roger Stafford le 11 Déc 2017
Write a loop to calculate the sum 1*2+2*3+3*4+ ... + 99*100. The answer is 333300
so far I have
total = 0;
for n = 1:100;
total = total + n;
end
fprintf('Total %d\n' , total)

Réponses (2)

John D'Errico
John D'Errico le 11 Déc 2017
Modifié(e) : John D'Errico le 11 Déc 2017
Your loop sums only the integers from 1 to n, not the product of consecutive integers. But, hey, you were close, and you made a credible effort.
total = 0;
for n = 1:99
total = total + n*(n+1);
end
fprintf('Total %d\n' , total)
There was actually a second problem with what you wrote. The desired sum will have only 99 terms to sum! So my loop goes only from 1 to 99.
  1 commentaire
Mohammed Safwat
Mohammed Safwat le 11 Déc 2017
Thanks John!

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Roger Stafford
Roger Stafford le 11 Déc 2017
Modifié(e) : Roger Stafford le 11 Déc 2017
An alternative formula which reduces the amount of computation would be:
n = 99;
total = n*(n+1)*(n+2)/3;
That is true for any positive integer n.

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