Given a natural number **n**, find the sum of sum-series of 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 = 5Output:35Explanation:

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

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 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

At the end, print the value stored in Total_sum.

## C++

`// 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

`// 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

`# 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#

`// 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` |

## Javascript

`<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>` |

**Output:**

35

**Time complexity**: O(N)**Efficient approach:**

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

where N is the natural number

**Proof of above formula:**

Lets assume N = 5

- Then sum is sum of all below elements in table, lets call this “
**result**”

1 | ||||

1 | 2 | |||

1 | 2 | 3 | ||

1 | 2 | 3 | 4 | |

1 | 2 | 3 | 4 | 5 |

lets populate the empty cells with same value in other columns, lets call this “**totalSum**“

1 | 2 | 3 | 4 | 5 |

1 | 2 | 3 | 4 | 5 |

1 | 2 | 3 | 4 | 5 |

1 | 2 | 3 | 4 | 5 |

1 | 2 | 3 | 4 | 5 |

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

- Since,

result = totalSum – populatedData

= N * [(N*(N+1))/2] – [(N-1) * (N) * (N+1)]/3

= (N*(N+1)*(N+2))/6

- Therefore

Below is the implementation of the above approach:

## C++

`// 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

`// 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

`# 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#

`// 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` |

## Javascript

`<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>` |

**Output:**

35

* Time complexity O(1)* considering multiplication, addition & division take constant time.

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