Solution 2590360

Submitted on 21 Jun 2020 by Piero Cimule

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Piero Cimule on 21 Jun 2020

C=[3,4,5];
for i=2:n
C(end+1,1,1)=C(end,2,1);
C(end,2,1)=C(end-1,3,1);
C(end,3,1)=sqrt(C(end,1,1)^2+C(end,2,1)^2);
end

C= [C(end,:,1)]
area=C(3)^2

Test	Status	Code Input and Output
1	Pass	n = 1; area_correct = 25; tolerance = 1e-12; assert(abs(triangle_sequence(n)-area_correct)<tolerance) C = 3 4 5 area = 25
2	Pass	n = 2; area_correct = 41; tolerance = 1e-12; assert(abs(triangle_sequence(n)-area_correct)<tolerance) C = 4.0000 5.0000 6.4031 area = 41
3	Pass	n = 3; area_correct = 66; tolerance = 1e-12; assert(abs(triangle_sequence(n)-area_correct)<tolerance) C = 5.0000 6.4031 8.1240 area = 66.0000
4	Pass	n = 50; area_correct = 439116598409; tolerance = 1e-3; assert(abs(triangle_sequence(n)-area_correct)<tolerance) C = 1.0e+05 * 4.0955 5.2095 6.6266 area = 4.3912e+11

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Problem 2024. Triangle sequence