Sum of sum of first n natural numbers

Given a positive integer n. The task is to find the sum of the sum of first n natural number.

Examples:

Input : n = 3
Output : 10
Sum of first natural number: 1
Sum of first and second natural number: 1 + 2 = 3
Sum of first, second and third natural number = 1 + 2 + 3 = 6
Sum of sum of first three natural number = 1 + 3 + 6 = 10

Input : n = 2
Output : 4

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to one by one add triangular numbers.

C++

 /* CPP program to find sum  series 1, 3, 6, 10, 15, 21... and then find its sum*/ #include using namespace std;    // Function to find the sum of series int seriesSum(int n) {     int sum = 0;     for (int i=1; i<=n; i++)        sum += i*(i+1)/2;     return sum; }    // Driver code int main() {     int n = 4;     cout << seriesSum(n);     return 0; }

Java

 // Java program to find sum // series 1, 3, 6, 10, 15, 21... // and then find its sum*/ import java.io.*;    class GFG {                // Function to find the sum of series     static int seriesSum(int n)     {         int sum = 0;         for (int i = 1; i <= n; i++)         sum += i * (i + 1) / 2;         return sum;     }        // Driver code     public static void main (String[] args)      {         int n = 4;         System.out.println(seriesSum(n));                } }    // This article is contributed by vt_m

Python3

 # Python3 program to find sum # series 1, 3, 6, 10, 15, 21... # and then find its sum.    # Function to find the sum of series def seriessum(n):            sum = 0     for i in range(1, n + 1):         sum += i * (i + 1) / 2     return sum        # Driver code n = 4 print(seriessum(n))    # This code is Contributed by Azkia Anam.

C#

 // C# program to find sum // series 1, 3, 6, 10, 15, 21... // and then find its sum*/ using System;    class GFG {        // Function to find the sum of series     static int seriesSum(int n)     {         int sum = 0;                    for (int i = 1; i <= n; i++)             sum += i * (i + 1) / 2;                        return sum;     }        // Driver code     public static void Main()     {         int n = 4;                    Console.WriteLine(seriesSum(n));     } }    // This article is contributed by vt_m.

PHP



Output:

20

Time complexity : O(n)

An efficient solution is to use direct formula n(n+1)(n+2)/6

Mathematically, we need to find, Σ ((i * (i + 1))/2), where 1 <= i <= n
So, lets solve this summation,

Sum = Σ ((i * (i + 1))/2), where 1 <= i <= n
= (1/2) * Σ (i * (i + 1))
= (1/2) * Σ (i2 + i)
= (1/2) * (Σ i2 + Σ i)

We know Σ i2 = n * (n + 1) * (2*n + 1) / 6 and
Σ i = n * ( n + 1) / 2.
Substituting the value, we get,
Sum = (1/2) * ((n * (n + 1) * (2*n + 1) / 6) + (n * ( n + 1) / 2))
= n * (n + 1)/2 [(2n + 1)/6 + 1/2]
= n * (n + 1) * (n + 2) / 6

Below is the implementation of the above approach:

C++

 /* CPP program to find sum  series 1, 3, 6, 10, 15, 21... and then find its sum*/ #include using namespace std;    // Function to find the sum of series int seriesSum(int n) {     return (n * (n + 1) * (n + 2)) / 6;  }    // Driver code int main() {     int n = 4;     cout << seriesSum(n);     return 0; }

Java

 // java program to find sum // series 1, 3, 6, 10, 15, 21... // and then find its sum import java.io.*;    class GFG  {     // Function to find the sum of series     static int seriesSum(int n)     {         return (n * (n + 1) * (n + 2)) / 6;      }       // Driver code     public static void main (String[] args) {                    int n = 4;         System.out.println( seriesSum(n));                } }    // This article is contributed by vt_m

Python3

 # Python 3 program to find sum # series 1, 3, 6, 10, 15, 21... # and then find its sum*/    # Function to find the sum of series def seriesSum(n):        return int((n * (n + 1) * (n + 2)) / 6)       # Driver code n = 4 print(seriesSum(n))    # This code is contributed by Smitha.

C#

 // C# program to find sum // series 1, 3, 6, 10, 15, 21... // and then find its sum using System;    class GFG {            // Function to find the sum of series     static int seriesSum(int n)     {         return (n * (n + 1) * (n + 2)) / 6;     }        // Driver code     public static void Main()     {            int n = 4;                    Console.WriteLine(seriesSum(n));     } }    // This code is contributed by vt_m.

PHP



Output:

20

Time complexity : O(1)

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