The series is: 1-2+3-4+5-6+7-8...N terms, we have to find out the sum up to Nth terms.

该序列是： 1-2 + 3-4 + 5-6 + 7-8 ... N个项 ，我们必须找出第N个项之和。

Solution:

解：

Let's analyse this problem,

让我们分析这个问题，

    If we want Sum of this series up to 2nd term then sum will be:
        1-2 =-1
    Up to 3rd term:
        1-2+3 =2
    Up to 4th term 
        1-2+3-4 = -2
    Up to 5th term:
        1-2+3-4+5= 3
        .
        .
        .

Now we can conclude that if we want sum up to an Odd term then we always get sum as ((N+1)/2) and if we want sum up to an Even term the sum is in the form of (-1*(N/2)).

现在我们可以得出结论，如果要对一个奇数项求和，则总和为((N + 1)/ 2)，如果要对一个偶数项求和，则总和为(-1 * (N / 2))。

Now Let's make logic for this using c programming,

现在，让我们使用C编程为此逻辑，

#include <stdio.h>

//function for creating the sum of the 
//series up to Nth term
int series_sum(int n)
{
    if (n % 2 == 0)
        return (-(n / 2));
    else
        return ((n + 1) / 2);
}

// main code
int main()
{
    int n;
    printf("Series:1-2+3-4+5-6+7-8.....N\n");
    printf("Want some up to N terms?\nEnter the N term:");
    scanf("%d", &n);
    
    printf("Sum is:%d", series_sum(n));
    
    return 0;
}

Output

输出量

Series:1-2+3-4+5-6+7-8.....N
Want some up to N terms?
Enter the N term:10
Sum is:-5

Read more...

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• C programs by categories

  C程序分类

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  系列程序的总和C

• Star/Asterisks Pattern programs in C

  C中的星号/星号模式程序

翻译自: https://www.includehelp.com/c-programs/calculate-the-sum-of-the-series-1-2-3-4-5-6-7-8-n-terms.aspx