# Sum of the series 1.2.3 + 2.3.4 + … + n(n+1)(n+2)

Find the sum up to n terms of the series: 1.2.3 + 2.3.4 + … + n(n+1)(n+2). In this 1.2.3 represent the first term and 2.3.4 represent the second term .

Examples :

```Input : 2
Output : 30
1.2.3 + 2.3.4 = 6 + 24 = 30

Input : 3
Output : 90
```

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

Simple Approach We run a loop for i = 1 to n, and find the sum of (i)*(i+1)*(i+2).
And at the end display the sum .

## C++

 `// CPP program to find sum of the series ` `// 1.2.3 + 2.3.4 + 3.4.5 + ... ` `#include ` `using` `namespace` `std; ` ` `  `int` `sumofseries(``int` `n) ` `{ ` `    ``int` `res = 0; ` `    ``for` `(``int` `i = 1; i <= n; i++)  ` `        ``res += (i) * (i + 1) * (i + 2);     ` `    ``return` `res; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``cout << sumofseries(3) << endl; ` `    ``return` `0; ` `} `

## Java

 `// Java program to find sum of the series ` `// 1.2.3 + 2.3.4 + 3.4.5 + ... ` `import` `java.io.*; ` `import` `java.math.*; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``static` `int` `sumofseries(``int` `n) ` `    ``{ ` `    ``int` `res = ``0``; ` `    ``for` `(``int` `i = ``1``; i <= n; i++)  ` `        ``res += (i) * (i + ``1``) * (i + ``2``);  ` `    ``return` `res; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``System.out.println(sumofseries(``3``)); ` `    ``} ` `} `

## Python3

 `# Python 3 program to find sum of the series ` `# 1.2.3 + 2.3.4 + 3.4.5 + ... ` ` `  `def` `sumofseries(n): ` ` `  `    ``res ``=` `0` `    ``for` `i ``in` `range``(``1``, n``+``1``):  ` `        ``res ``+``=` `(i) ``*` `(i ``+` `1``) ``*` `(i ``+` `2``)  ` `    ``return` `res ` ` `  `# Driver Program ` `print``(sumofseries(``3``))  ` ` `  `# This code is contributed ` `# by Smitha Dinesh Semwal `

## C#

 `// Java program to find sum of the series ` `// 1.2.3 + 2.3.4 + 3.4.5 + ... ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``static` `int` `sumofseries(``int` `n) ` `    ``{ ` `        ``int` `res = 0; ` `        ``for` `(``int` `i = 1; i <= n; i++)  ` `            ``res += (i) * (i + 1) * (i + 2);  ` `        ``return` `res; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``Console.WriteLine(sumofseries(3)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` `

Output :

```90
```
```Complexity : O(N)
```

## Efficient Approach

Using Efficient Approach we know that we have to find = summation of( (n)*(n+1)*(n+2) )

Sn = summation[ (n)*(n+1)*(n+2) ]
Sn = summation [n3 + 2*n2 + n2 + 2*n]

We know sum of cubes of natural numbers is (n*(n+1))/2)2, sum of squares of natural numbers is n * (n + 1) * (2n + 1) / 6 and sum of first n natural numbers is n(n+1)/2

Sn = ((n*(n+1))/2)2 + 3((n)*(n+1)*(2*n+1)/6) + 2*((n)*(n+1)/2)
So by evaluating the above we get,
Sn = (n*(n+1)*(n+2)*(n+3)/4)
Hence it has a O(1) complexity.

## C++

 `// Efficient CPP program to  ` `// find sum of the series  ` `// 1.2.3 + 2.3.4 + 3.4.5 + ... ` `#include ` `using` `namespace` `std; ` ` `  `// function to calculate ` `// sum of series ` `int` `sumofseries(``int` `n) ` `{ ` `    ``return` `(n * (n + 1) *  ` `           ``(n + 2) * (n + 3) / 4); ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``cout << sumofseries(3) << endl; ` `    ``return` `0; ` `} `

## Java

 `// Efficient Java program to ` `// find sum of the series  ` `// 1.2.3 + 2.3.4 + 3.4.5 + .. ` `import` `java.io.*; ` `import` `java.math.*; ` ` `  `class` `GFG  ` `{ ` `    ``static` `int` `sumofseries(``int` `n) ` `    ``{ ` `    ``return` `(n * (n + ``1``) *  ` `           ``(n + ``2``) * (n + ``3``) / ``4``); ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``System.out.println(sumofseries(``3``)); ` `    ``}  ` `} `

## Python3

 `# Efficient CPP program to find sum of the ` `# series 1.2.3 + 2.3.4 + 3.4.5 + ... ` ` `  `# function to calculate sum of series ` `def` `sumofseries(n): ` ` `  `    ``return` `int``(n ``*` `(n ``+` `1``) ``*` `(n ``+` `2``) ``*` `(n ``+` `3``) ``/` `4``) ` ` `  ` `  `# Driver program ` `print``(sumofseries(``3``)) ` `     `  ` `  `# This code is contributed ` `# by Smitha Dinesh Semwal `

## C#

 `// Efficient C# program to  ` `// find sum of the series  ` `// 1.2.3 + 2.3.4 + 3.4.5 + .. ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `    ``static` `int` `sumofseries(``int` `n) ` `    ``{ ` `    ``return` `(n * (n + 1) *  ` `           ``(n + 2) * (n + 3) / 4); ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``Console.WriteLine(sumofseries(3)); ` `    ``}  ` `} ` ` `  `// This code is contributed by anuj_67. `

## PHP

 ` `

Output :

```90
```

Time Complexity : O(1)

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : vt_m

Article Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.