Calculating the sum in C++

Question

How do I calculate the sum to below in C++? I tried the following code but failed.

enter image description here

#include <iostream>
using namespace std;
int main()
{
    int n, p, r = -1;
    cin >> n;
    for (p = 0; p < 10; p++)
        r *= (-1);
    cout << r << endl;
    return 0;
}

Why are you inputting n? Your expression says that n starts from 0 and goes to 10. — Phonon
– Phonon, Commented Jan 26, 2012 at 16:47
I suspect the sum should read sum( n=0, 10, (-1)^(n+1) ), so the result is either -1 or 0 — Loebl
– Loebl, Commented Jan 26, 2012 at 16:48
that's still basic maths, no need for a loop, you just -1 * upperLimit%2 — Flavius
– Flavius, Commented Jan 26, 2012 at 16:51

warren · Accepted Answer · 2012-01-26 17:54:48Z

4

#include <iostream>
using namespace std;
int main()
{
    int p, r = 1;
    int iSum=0;
    // cin >> n;

    for (p = 0; p <= 10; p++)
    {
        r *= (-1);
        iSum+=r;
    }
    cout << iSum << endl;
    return 0;
}

edited Jan 26, 2012 at 17:54

warren

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answered Jan 26, 2012 at 16:46

MRM

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Comments

Flavius · Accepted Answer · 2012-01-27 00:42:57Z

4

That's basic mathematics, there is no need for a loop, you can just calculate -1 * (upperLimit + 1)%2.

Look at the series and think: -1 +1 -1 +1 -1 +1 ....

edited Jan 27, 2012 at 0:42

answered Jan 26, 2012 at 16:52

Flavius

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3 Comments

Jonathan Henson Over a year ago

+1 yes, in general it would be x * upperLimit % 2. Great answer!

Jonathan Henson Over a year ago

@MRM, that would produce a positive 1 or a 0, depending on the upderLimit. The only possible sums are -1 or 0. 10 % 2 = 0, 0 * -1 = 0. 11 % 2 = 1, 1 * -1 = -1. I think this is in an old Calculus book of mine as a telescoping series. I'll verify.

Mooing Duck Over a year ago

While accurate, this doesn't apply to more complex examples.

prajmus · Accepted Answer · 2012-01-26 18:22:05Z

1

Although @Flavius is right, the sum starts from 0 so it'll be -1 * (upperLimit+1)%2 as the sum iterates not 10 but 11 times. The upperLimit%2 thing works for sums starting at 1

P.S: Sorry for answering, I cannot yet comment, just registered.

answered Jan 26, 2012 at 18:22

prajmus

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1 Comment

Flavius Over a year ago

Thanks, I was aware of that but I was too lazy to fix it. +1 and welcome on SO.

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