# Sum of the series 1, 3, 6, 10… (Triangular Numbers)

• Difficulty Level : Medium
• Last Updated : 15 Jun, 2022

Given n, no of elements in the series, find the summation of the series 1, 3, 6, 10….n. The series mainly represents triangular numbers.
Examples:

```Input: 2
Output: 4
Explanation: 1 + 3 = 4

Input: 4
Output: 20
Explanation: 1 + 3 + 6 + 10 = 20```

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

 ``

## Javascript

 ``

Output:

`20`

Time complexity : O(n)

Auxiliary Space: O(1) since using constant variables

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

```Let g(i) be i-th triangular number.
g(1) = 1
g(2) = 3
g(3) = 6
g(n) = n(n+1)/2```

```Let f(n) be the sum of the triangular
numbers 1 through n.
f(n) = g(1) + g(2) + ... + g(n)

Then:
f(n) = n(n+1)(n+2)/6```

How can we prove this? We can prove it by induction. That is, prove two things :

1. It’s true for some n (n = 1, in this case).
2. If it’s true for n, then it’s true for n+1.

This allows us to conclude that it’s true for all n >= 1.

```Now 1) is easy. We know that f(1) = g(1)
= 1. So it's true for n = 1.

Now for 2). Suppose it's true for n.
Consider f(n+1). We have:
f(n+1) = g(1) + g(2) + ... + g(n) + g(n+1)
= f(n) + g(n+1)

Using our assumption f(n) = n(n+1)(n+2)/6
and g(n+1) = (n+1)(n+2)/2, we have:
f(n+1) = n(n+1)(n+2)/6 + (n+1)(n+2)/2
= n(n+1)(n+2)/6 + 3(n+1)(n+2)/6
= (n+1)(n+2)(n+3)/6
Therefore, f(n) = 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

 ``

## Javascript

 ``

Output:

`20`

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

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