Find the sum of the series 1+(1+2)/2+(1+2+3)/3+... till N terms

Given a number N, the task is to find the sum of the below series till N terms.

1+(1+2)/2+(1+2+3)/3+... till N terms

Examples:

Input: N = 3 
Output: 4.5

Input: N = 4
Output: 7

Approach:  

From the given series, find the formula for the Nth term:

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

Derivation:

The following series of steps can be used to derive the formula to find the sum of N terms-

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

Illustration:

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

Below is the C++ program to implement the above approach:

// C++ program to find the sum of the
// series 1+(1+2)/2+(1+2+3)/3+...
// till N terms
#include <bits/stdc++.h>
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 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 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# 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.

  <script>
        // 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
    </script>

13.5

Time Complexity: O(1) 
Auxiliary Space: O(1), since no extra space has been taken.