求系列 1+(1+2)/2+(1+2+3)/3+... 之和，直到 N 项

给定一个数字N ，任务是找到以下系列的总和，直到N项。

1+(1+2)/2+(1+2+3)/3+… till N terms

例子：

Input: N = 3 
Output: 4.5

Input: N = 4
Output: 7

方法：

从给定的系列中，找到第 N项的公式：

1st term = 1 = 1

2nd term = (1+2)/2 = 1.5

3rd term = (1+2+3)/3 = 2

4th term = (1+2+3+4)/4 = 2.5

.

.

Nth term = N*(N +1)/(2*N) = (N+1)/2

推导：

以下一系列步骤可用于推导公式以找到 N 项的总和 -

The sequence-

1+(1+2)/2+(1+2+3)/3+… till N terms

can be written as-

1 +1.5 +2 +2.5 +3 +3.5 +4 +…till N terms

The above series is an Arithmetic Progression(AP) series. So, we can directly apply the formula of sum of N terms in AP.

The Sum of N terms of an AP can also be given by S_N,

S_N = N * [First term + Last term] / 2

S_N = N * [2 * a + (N – 1) * d] / 2        -(1)

From the above equation, it is known that,  

a(first term)=1, d(common difference) = 1.5 -1= 0.5

Substituting the values of a and d in the equation (1), we get-

S_N = N * (2 + (N – 1) * 0.5) / 2

插图：

Input: N = 5
Output: 10
Explanation:
S_N = 5 * [2 * 1 + (5 – 1) * 0.5] / 2
     = 5 * (2 + 2) / 2
     = 5 * 2
     = 10

下面是实现上述方法的 C++ 程序：

C++

// C++ program to find the sum of the
// series 1+(1+2)/2+(1+2+3)/3+...
// till N terms
#include 
using namespace std;
 
// Function to return the sum
// upto N term of the series
double sumOfSeries(int N)
{
    return ((double)N
            * (2 + ((double)N - 1) * 0.5))
           / 2;
}
 
// Driver Code
int main()
{
    // Get the value of N
    int N = 6;
 
    cout << sumOfSeries(N);
    return 0;
}

Java

// Java program to find the sum of the
// series 1+(1+2)/2+(1+2+3)/3+...
// till N terms
import java.util.*;
public class GFG
{
 
  // Function to return the sum
  // upto N term of the series
  static double sumOfSeries(int N)
  {
    return ((double)N
            * (2 + ((double)N - 1) * 0.5))
      / 2;
  }
 
  // Driver Code
  public static void main(String args[])
  {
 
    // Get the value of N
    int N = 6;
 
    System.out.println(sumOfSeries(N));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Python

# Python program to find the sum of the
# series 1+(1+2)/2+(1+2+3)/3+...
# till N terms
 
# Function to return the sum
# upto N term of the series
def sumOfSeries(N):
     
    return (N * (2 + (N - 1) * 0.5) / 2)
 
# Driver Code
# Get the value of N
N = 6
print(sumOfSeries(N))
 
# This code is contributed by Samim Hossain Mondal.

C#

// C# program to find the sum of the
// series 1+(1+2)/2+(1+2+3)/3+...
// till N terms
using System;
class GFG
{
 
  // Function to return the sum
  // upto N term of the series
  static double sumOfSeries(int N)
  {
    return ((double)N
            * (2 + ((double)N - 1) * 0.5))
      / 2;
  }
 
  // Driver Code
  public static void Main()
  {
 
    // Get the value of N
    int N = 6;
 
    Console.Write(sumOfSeries(N));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

      // JavaScript code for the above approach
 
      // Function to return the sum
      // upto N term of the series
      function sumOfSeries(N) {
          return (N
              * (2 + (N - 1) * 0.5))
              / 2;
      }
 
      // Driver Code
 
      // Get the value of N
      let N = 6;
      document.write(sumOfSeries(N));
 
       // This code is contributed by Potta Lokesh
  

输出

13.5

时间复杂度： O(1)
辅助空间： O(1)