While loop for sums?

Add up the squares of all odd positive integers until it equals or exceeds 5 million.

(1^2+3^2...)

[Steven Lord]

Show what you've tried to do to solve the problem and ask a specific question about where you're having difficulty and you may receive some guidance.

[Rena Berman]

(Answers Dev) Restored Question.

Answer 1

[Image Analyst]

Try this:

theSum = 0; % Initialize
thisNumber = -1;
while theSum < 5000000
    thisNumber = thisNumber + .......
    theSum = theSum + ......
end

I've given you a start. Please finish the rest of your homework yourself.

Answer 2

[s.p4m]

sum=0;
k=0;
while(sum<=5*10^6)
  if(mod(k,2))
    sum=sum+k^2;
  end
  k=k+1;
end

Next time please try to solve your homework by yourself

[Image Analyst]

I see 3 problems with this.

[s.p4m]

I found one mistake, where I use i instead of k in the mod() function, but everything else seems to be fine.

[Image Analyst]

OK, 4 problems then:

1. Using i (imaginary variable) for loop index, like you said.
2. Using <= when the user said to quit when the sum hit 5 million. Using <= will do another iteration after it should have stopped. Actually it's not really a problem since 5 million is not an odd integer so it won't increase the sum, but it is another unneeded iteration.
3. Using sum, a built in function name, as the name of your variable will prevent you from using the sum function in the same scope. It's never a good idea to blow away/overwrite built-in functions.
4. Iterating on every single integer will do twice as many iterations as my solution where I hope the poster figured out that you're supposed to add 2 to the loop index. Plus mod() will slow it down a little bit. It's unnecessary - just add 2 to get only odd integers.

And of course there is the issue of just answering a homework question outright. Our protocol here is not to do homework for people so they can turn in our solution as their own, but to give them hints or starter snippets that they can modify so they at least have some ownership of the solution.

[s.p4m]

Thanks for the answer. You are right with every point.

I didn't know about the rule not to do outher people homework, but I will embrace it from now on.

[Jan]

5. 5*10^6 is an expensive power operation, while 5e6 is cost free constant.

Thanks, s.p4m, for you suggestion. If the OP reads the comments carefully, he has learned something about programming. :-)

Answer 3

[prin]

jum=100; n = 1; while sum(1:r) > jum

disp(r)

n = n - 1;

[Walter Roberson]

r is not defined. You are missing the "end" of the "while".