Thread: need help with series-- please

I would suggest using a for loop.

Ignore the man before me. The while loop is superior in every way.

Originally Posted by ggs

Ignore the man before me. The while loop is superior in every way.

Considering the for loop is a while loop condensed into one statement what you said doesnt really make any sense. And Salem is a very seasoned proffesional so he was is correct 99% of the time. Although either loop would work well.

Well you want 20 terms, so how about
for ( nTerms = 0 ; nTerms < 20 ; nTerms++ )

Don't think we're being mean or anything, but it's hard to describe what is basically 1 line of code without giving the whole game away.

Post an effort of your own.

The first term is 4 (or 4/1).
The second term is -4/3.
The third term is 4/5.
The fourth term is -4/7.

Do you see what is happening? The denominator increases by 2 every term and the sign flips every term. The sign flip can be done with a multiplication of -1 and the other, well... if you don't know how to add 2 to something you're in the wrong business. The rest is straight addition of those terms.

Now now. "Explain to me more." Consider what it is you are having trouble understanding (this might take some thought, maybe a snack or a quick nap), and then ask a question. Use messages to contact board members, not posts.

Anyone who read hk's post knows how to do it. Except you.

To start off, you need something to hold the sum, you need something to hold the current value of the denominator, and you need something to store the current sign. Taking all of that gets us:

Code:

double sum = 0.0;
double denominator = 1.0;
double sign = 1.0;

Now you also need your loop:

Code:

for( int i = 0; i < 20; ++i )
{
    //  What goes here?
}

In the body of the loop you need to increment the sum by the current term (sign multiplied by 4.0 divided by the value of the denominator). Then, you need to increment the denominator by 2, and multiply the sign by -1 so that it flips from positive to negative on each loop iteration. You also need to print the current value of the sum.