Squares of all positive odd integers whose sum is greater than or equal to 3,000,000

I'm trying to create a program that calculates the squares of all positive odd integers whose sum is greater than or equal to 3,000,000. The problem is that I'm not given a specified number of terms to use and was wondering if there was any way to do this without just specifying the amount of terms by hand.

array = [];
sum = 0;
n = 0;
while sum <= 3000000
   for i = 1:2:n+2
       array(i) = i^2;
       sum = sum + i^2;
   end
   n = n + 1;
end 

I was trying something like this but i'm pretty sure my variable n is what's messing up the for loop from giving me the right increments.

[James Tursa]

Can you post the actual wording of the assignment so we can be sure what the algorithm is supposed to do? I can't quite tell from how you have worded this what is actually required.

[Noah Sickels]

Sorry about that.

"Suppose that the squares of all positive odd integers were to be added up until this sum first equaled or exceeded 3 million. How many terms would be added? What is the final sum?"

[Bjorn Gustavsson]

It is not explicitly specified that you need to start at the smallest positive odd integer, is it?

[Noah Sickels]

It's not specified but i presummed it would start at 1 and go from there. He's not very specific on any of these projects which makes the actual program hard to complete in the way he desires.

[Steven Lord]

First, you shouldn't use sum as a variable name. While that variable exists you won't be able to call the function named sum.

Second, there's no need for a for loop in this assignment. The while keyword is a looping construct. As the old folk song goes:

bottlesOfBeerOnTheWall = 99;
while bottlesOfBeerOnTheWall > 0
    fprintf("%d bottles of beer on the wall\n", bottlesOfBeerOnTheWall)
    fprintf("%d bottles of beer\n", bottlesOfBeerOnTheWall)
    fprintf("You take one down\n");
    bottlesOfBeerOnTheWall = bottlesOfBeerOnTheWall-1;
    fprintf("You pass it around\n");
    fprintf("%d bottles of beer on the wall\n\n", bottlesOfBeerOnTheWall)
end

This will execute the five fprintf statements and the one subtraction until the condition is no longer satisfied. I think you've already identified the equivalent of putting the bottles of beer on the wall and checking if there is any beer on the wall. What's the equivalent of taking the nth bottle of beer from the wall and passing it around for your assignment?

Answer 1

[Sunny Choudhary]

Hi

I debugged your code by printing i values.

What mistake I can see in your code is its printing sum of squares of 1 to 89 for multiple times

You are doing = (1^2 + 2^2+ ... + 89^2) some k times

But what we want is 1^2 + 2^2+ ... + 132^2) for single time

You can use this code to calculate the squares of all positive odd integers whose sum is greater than or equal to 3,000,000.

sum = 0;
n = 0;
i = 0;
while sum < 3000000
   sum = sum + i * i;
   n = n + 1;
   i = i + 2;
end 
n

n = 132

sum

sum = 3031864