What is the value of the variable, p, at the end of this for loop?

p = 0;

for k = 1:3:9

p = p + k;

end;

Question

What is the value of the variable, p, at the end of this for loop?

p = 0;

for k = 1:3:9

p = p + k;

end;

BlackTom AI · Accepted Answer

To solve this, I’ll examine how the for loop operates and what values k takes.
First, interpret the loop syntax p = 0; for k = 1:3:9; p = p + k; end; In MATLAB/Octave, 1:3:9 generates a sequence starting at 1, incrementing by 3, up to 9, inclusive if possible. This yields the values k = 1, 4, 7; the value 10 would exceed 9 and is therefore not included.
Next, track p through each iteration: initial p = 0. When k = 1, p becomes 0 + 1 = 1. When k = 4, p becomes 1 + 4 = 5. When k = 7, p becomes 5 + 7 = 12. There are no more iterations since the next candidate would be 10 which is greater than 9.
Now evaluate the given options:
- Option 9: This would imply only the first value (or an incorrect sum) was considered; however, the actual loop includes 1, 4, and 7, so 9 is too small.
- Option 12: This matches the computed total 1 + 4 + 7, after summing all loop iterations.
- Option 21: This would correspond to summing 1 + 4 + 7 + 9, but 9 is not part of the sequence generated by 1:3:9, since 9 would require a step of 2 in this sequence; thus 21 is incorrect.
- Option 11: This would reflect an off-by-one or a partial sum error (e.g., stopping early or miscounting one term). The actual terms are 1, 4, and 7, which sums to 12, not 11.
In summary, understanding the exact sequence produced by 1:3:9 is key: it does not include 9 itself, and the loop accumulates 1 + 4 + 7 = 12. The other options arise from miscalculating the sequence bounds or the cumulative sum, leading to values 9, 21, or 11 instead of the correct total.

ENGR10004_2025_SM1 Homework Assessment 9 - Image Processing Quiz

What is the value of the variable, p, at the end of this for loop? p = 0; for k = 1:3:9 p = p + k; end;

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