求出1/2 – 2/3 + 3/4 – 4/5 +…的序列之和，直到N项

给定数字N ，任务是找到以下序列的和，直到N个项。

$\frac{1}{2} - \frac{2}{3} + \frac{3}{4} - \frac{4}{5} + ...$

例子：

Input: N = 6 
Output: -0.240476
Input: N = 10 
Output: -0.263456 
 

方法：从给定的系列中，找到第N个项的公式：

1st term = 1/2
2nd term = - 2/3
3rd term = 3/4
4th term = - 4/5
.
.
Nthe term = ((-1)N) * (N / (N + 1))

所以：

Nth term of the series 

然后对[1，N]范围内的数字进行迭代，以使用上述公式查找所有项并计算其总和。
下面是上述方法的实现：

C++

*** QuickLaTeX cannot compile formula:
 

*** Error message:
Error: Nothing to show, formula is empty

Java

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the sum of series
void printSeriesSum(int N)
{
    double sum = 0;
 
    for (int i = 1; i <= N; i++) {
 
        // Generate the ith term and
        // add it to the sum if i is
        // even and subtract if i is
        // odd
        if (i & 1) {
            sum += (double)i / (i + 1);
        }
        else {
            sum -= (double)i / (i + 1);
        }
    }
 
    // Print the sum
    cout << sum << endl;
}
 
// Driver Code
int main()
{
    int N = 10;
 
    printSeriesSum(N);
    return 0;
}

Python3

// Java program for the above approach
class GFG{
  
// Function to find the sum of series
static void printSeriesSum(int N)
{
    double sum = 0;
  
    for (int i = 1; i <= N; i++) {
  
        // Generate the ith term and
        // add it to the sum if i is
        // even and subtract if i is
        // odd
        if (i % 2 == 1) {
            sum += (double)i / (i + 1);
        }
        else {
            sum -= (double)i / (i + 1);
        }
    }
  
    // Print the sum
    System.out.print(sum +"\n");
}
  
// Driver Code
public static void main(String[] args)
{
    int N = 10;
  
    printSeriesSum(N);
}
}
 
// This code is contributed by 29AjayKumar

C#

# Python3 program for the above approach
 
# Function to find the sum of series
def printSeriesSum(N) :
     
    sum = 0;
 
    for i in range(1, N + 1) :
 
        # Generate the ith term and
        # add it to the sum if i is
        # even and subtract if i is
        # odd
        if (i & 1) :
            sum += i / (i + 1);
      
        else :
            sum -= i / (i + 1);
     
 
    # Print the sum
    print(sum);
 
# Driver Code
if __name__ == "__main__" :
 
    N = 10;
 
    printSeriesSum(N);
    
    # This code is contributed by Yash_R

Javascript

// C# program for the above approach
using System;
  
class GFG {
      
// Function to find the sum of series
static void printSeriesSum(int N)
{
    double sum = 0;
 
    for (int i = 1; i <= N; i++) {
 
        // Generate the ith term and
        // add it to the sum if i is
        // even and subtract if i is
        // odd
        if ((i & 1)==0) {
            sum += (double)i / (i + 1);
        }
        else {
            sum -= (double)i / (i + 1);
        }
    }
 
    // Print the sum
    Console.WriteLine(sum);
}
 
// Driver Code
    public static void Main (string[] args)
    {
        
    int N = 10;
 
    printSeriesSum(N);
}
}
 
// This code is contributed by shivanisinghss2110

输出：

 
// javascript program for the above approach
 
// Function to find the sum of series
function printSeriesSum( N)
{
    let sum = 0;
 
    for (let i = 1; i <= N; i++) {
 
        // Generate the ith term and
        // add it to the sum if i is
        // even and subtract if i is
        // odd
        if (i & 1) {
            sum += i / (i + 1);
        }
        else {
            sum -= i / (i + 1);
        }
    }
 
    // Print the sum
     document.write( sum.toFixed(6) );
}
 
// Driver Code
 
    let N = 10;
 
    printSeriesSum(N);
    
// This code is contributed by todaysgaurav