Sum of sum-series of first N Natural numbers

Given a natural number n, find the sum of the sum-series of the first N natural number.

Sum-Series: is sum of first N natural numbers, i.e, sum-series of 5 is 15 ( 1 + 2 + 3 + 4 + 5 ).
 

Natural number	1	2	3	4	5	6
Sum of natural number (sum-series)	1	3	6	10	15	21
Sum of sum-series	1	4	10	20	35	56

Example: 

Input: N = 5 
Output: 35 
Explanation: 
Sum of sum-series of {1, 2, 3, 4, 5} i.e. {1 + 3 + 6 + 10 + 15} is 35.

Input: N = 2 
Output: 4 
Explanation: 
Sum of sum-series of {1, 2} i.e. {1 + 3} is 4. 

Simple approach: 
Find sum series for every value from 1 to N and then add it. 

• Create a variable Total_sum to store the required sum series.
• Iterate over the number from 1 to N
• Find sum-series of every value by using the formulae sum = (N*(N + 1)) / 2
• Add the value to Total_sum
• In the end, print the value stored in Total_sum. 

Below is the implementation of the above approach:

// C++ program to implement
// the above approach
#include<bits/stdc++.h>
using namespace std;

// Function to find the sum
static long sumOfSumSeries(int N)
{
    long sum = 0L;

    // Calculate sum-series
    // for every natural number
    // and add them
    for (int i = 1; i <= N; i++) 
    {
        sum = sum + (i * (i + 1)) / 2;
    }

    return sum;
}

// Driver code
int main()
{
    int N = 5;
    cout << sumOfSumSeries(N);
}

// This code is contributed by Code_Mech

// Java program to implement
// the above approach

class GFG {

    // Function to find the sum
    static long sumOfSumSeries(int N)
    {

        long sum = 0L;

        // Calculate sum-series
        // for every natural number
        // and add them
        for (int i = 1; i <= N; i++) {
            sum = sum + (i * (i + 1)) / 2;
        }

        return sum;
    }

    // Driver code
    public static void main(String[] args)
    {
        int N = 5;
        System.out.println(sumOfSumSeries(N));
    }
}

# Python3 program to implement 
# the above approach 

# Function to find the sum 
def sumOfSumSeries(N): 

    _sum = 0

    # Calculate sum-series 
    # for every natural number 
    # and add them 
    for i in range(N + 1): 
        _sum = _sum + (i * (i + 1)) // 2

    return _sum 

# Driver code 
N = 5

print(sumOfSumSeries(N)) 
    
# This code is contributed by divyamohan123 

// C# program to implement
// the above approach
using System;
class GFG{

// Function to find the sum
static long sumOfSumSeries(int N)
{
    long sum = 0L;

    // Calculate sum-series
    // for every natural number
    // and add them
    for(int i = 1; i <= N; i++) 
    {
       sum = sum + (i * (i + 1)) / 2;
    }
    
    return sum;
}

// Driver code
public static void Main()
{
    int N = 5;
    
    Console.Write(sumOfSumSeries(N));
}
}

// This code is contributed by Nidhi_Biet

<script>

    // Javascript program to implement
    // the above approach
    
    // Function to find the sum
    function sumOfSumSeries(N)
    {
        let sum = 0;

        // Calculate sum-series
        // for every natural number
        // and add them
        for (let i = 1; i <= N; i++) 
        {
            sum = sum + (i * (i + 1)) / 2;
        }

        return sum;
    }
    
    let N = 5;
    document.write(sumOfSumSeries(N));
    
    // This code is contributed by suresh07.
</script>

35

Time complexity: O(N)
Auxiliary Space: O(1)

Efficient approach: 
Total_sum of the above series can be calculated directly by using the below formulae: 

\text{Total sum} = \frac{(N*(N+1)*(N+2))}{6}      
where N is the natural number 

Proof of the above formula:
Lets assume N = 5 

1. Then the sum is sum of all the below elements in the table, let's call this "result" 
    

let's populate the empty cells with the same value in other columns, lets's call this "totalSum"

As sum of N numbers is repeated N times 
totalSum = N * [(N*(N + 1))/2]
populated data = (1 times * 2) + (2 times * 3) + (3 times * 4) + (4 times * 5) 
= 1*2 + 2*3 + 3*4 ......... +(N-1)*N 
=[(N-1) * (N) * (N+1)]/3 
 

1. Since, 
    

result = totalSum - populatedData 
= N * [(N*(N+1))/2] - [(N-1) * (N) * (N+1)]/3 
= (N*(N+1)*(N+2))/6  

1. Therefore 
   \text{Sum of Sum-Series till N} = \frac{(N*(N+1)*(N+2))}{6}      
    

Below is the implementation of the above approach: 

// C++ program to implement
// the above approach
#include <iostream>
#include <math.h>
using namespace std;

// Function to find the sum
long sumOfSumSeries(int n)
{
    return (n * (n + 1) * (n + 2)) / 6;
}

// Driver code
int main ()
{
    int N = 5;
    cout << sumOfSumSeries(N);
    return 0;
}

// This code is contributed
// by shivanisinghss2110

// Java program to implement
// the above approach

class GFG {

    // Function to find the sum
    static long sumOfSumSeries(int n)
    {
        return (n * (n + 1) * (n + 2)) / 6;
    }

    // Driver code
    public static void main(String[] args)
    {
        int N = 5;
        System.out.println(sumOfSumSeries(N));
    }
}

# Python3 program to implement 
# the above approach

# Function to find the sum 
def sumOfSumSeries(n): 
    
    return (n * (n + 1) * (n + 2)) // 6

# Driver code 
N = 5

print(sumOfSumSeries(N))

# This code is contributed by divyamohan123 

// C# program to implement the 
// above approach
using System;
class GFG{

// Function to find the sum
static long sumOfSumSeries(int n)
{
    return (n * (n + 1) * (n + 2)) / 6;
}

// Driver code
public static void Main(String[] args)
{
    int N = 5;
    
    Console.Write(sumOfSumSeries(N));
}
}

// This code is contributed by Ritik Bansal

<script>
    // Javascript program to implement
    // the above approach
    
    // Function to find the sum
    function sumOfSumSeries(n)
    {
        return (n * (n + 1) * (n + 2)) / 6;
    }
    
    let N = 5;
    document.write(sumOfSumSeries(N));

</script>

35

Time complexity: O(1), considering multiplication, addition & division takes constant time.
Auxiliary Space: O(1)