How would you write this series into a for loop?

Question

$$\sum_{i=1}^5 (-1)^{i+1}*i^2 $$ I thought it would be something like this but i always get 36 or -36. This is c also.

for(i = 1; i <= n; i++)
    if(i++ % 2 == 0) {
        count = pow(-1,n+1)*pow(n,2);
    } else{
        count = pow(-1,n+1)*pow(n,2);
    }

I didnt see that, i removed it. – billnye852 Feb 23 '17 at 03:05 — billnye852, Feb 23 '17 at 03:05
It would be better to ask such questions on stackoverflow – Feb 23 '17 at 08:54 — , Feb 23 '17 at 08:54

score 2 · Accepted Answer · answered Feb 23 '17 at 03:08

2

This should suffice

int count = 0;
for(i=1;i<=n;i++)
{
    count += pow(-1,i+1)*i*i;
}

n is just the stop condition, it doesn't need to be referenced explicitly in the loop.

Also you can probably see that you don't need to break the addition into conditions since the arithmetic is the same for both.

answered Feb 23 '17 at 03:08

kojak

Omg thank you soo much, ive been at it for the entire day and i didnt was just stuck. – billnye852 Feb 23 '17 at 03:14

score 0 · Answer 2 · answered Feb 23 '17 at 03:06

0

First thing you have to find sum for 5 terms. So loop should be from n=1 to 5.

Second here is no need of even odd concept.

answered Feb 23 '17 at 03:06

Amar

score 0 · Answer 3 · answered Feb 23 '17 at 03:06

$2$ things I would like to point out:

First of all, the if statement will be
if(i%2 == 0)
Secondly, the body of both the if and else statements should contain $\mathrm{pow(i,2)}$ rather than $\mathrm{pow(n,2)}$.
Thirdly, the body of both the if and else statements should also contain $\mathrm{count +=}$ rather than $\mathrm{count=}$

Hope this helps.

3 Answers3