# Program to find sum of series 1*2*3 + 2*3*4+ 3*4*5 + . . . + n*(n+1)*(n+2)

Last Updated : 29 Mar, 2023

Given a positive integer n and the task is to find the sum of series 1*2*3 + 2*3*4 + 4*5*6 + . . .+ n*(n+1)*(n+2).
Examples:

```Input : n = 10
Output : 4290
1*2*3 + 2*3*4 + 3*4*5 + 4*5*6 + 5*6*7 + 6*7*8 +
7*8*9 + 8*9*10 + 9*10*11 + 10*11*12
= 6 + 24 + 60 + 120 + 210 + 336 + 504 +
720 + 990 + 1320
= 4290

Input : n = 7
Output : 1260```

Method 1: In this case loop will run n times and calculate the sum.

## C++

 `// Program to find the sum of series` `// 1*2*3 + 2*3*4 + . . . + n*(n+1)*(n+1)` `#include ` `using` `namespace` `std;` `// Function to calculate sum of series.` `int` `sumOfSeries(``int` `n)` `{` `    ``int` `sum = 0;` `    ``for` `(``int` `i = 1; i <= n; i++)` `        ``sum = sum + i * (i + 1) * (i + 2);` `    ``return` `sum;` `}`   `// Driver function` `int` `main()` `{` `    ``int` `n = 10;` `    ``cout << sumOfSeries(n);` `    ``return` `0;` `}`

## Java

 `// Java Program to find the sum of series` `// 1*2*3 + 2*3*4 + . . . + n*(n+1)*(n+1)` `public` `class` `GfG{`   `    ``// Function to calculate sum of series.` `    ``static` `int` `sumOfSeries(``int` `n)` `    ``{` `        ``int` `sum = ``0``;`   `        ``for` `(``int` `i = ``1``; i <= n; i++)` `            ``sum = sum + i * (i + ``1``) * (i + ``2``);`   `        ``return` `sum;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String s[])` `    ``{` `        ``int` `n = ``10``;` `        ``System.out.println(sumOfSeries(n));` `    ``}` `} `   `// `

## Python3

 `# Python program to find the` `# sum of series` `# 1*2*3 + 2*3*4 + . . .` `# + n*(n+1)*(n+1)`   `# Function to calculate sum` `# of series.` `def` `sumOfSeries(n):` `    ``sum` `=` `0``;` `    ``i ``=` `1``;` `    ``while` `i<``=``n:` `        ``sum` `=` `sum` `+` `i ``*` `(i ``+` `1``) ``*` `(` `                                ``i ``+` `2``)` `        ``i ``=` `i ``+` `1` `    ``return` `sum`   `# Driver code` `n ``=` `10` `print``(sumOfSeries(n))`   `# This code is contributed by "Abhishek Sharma 44"`

## C#

 `// C# Program to find the sum of series` `// 1*2*3 + 2*3*4 + . . . + n*(n+1)*(n+1)` `using` `System;`   `public` `class` `GfG` `{`   `    ``// Function to calculate sum of series.` `    ``static` `int` `sumOfSeries(``int` `n)` `    ``{` `        ``int` `sum = 0;`   `        ``for` `(``int` `i = 1; i <= n; i++)` `            ``sum = sum + i * (i + 1) * (i + 2);`   `        ``return` `sum;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `n = 10;` `        ``Console.WriteLine(sumOfSeries(n));` `    ``}` `} `   `// This article is contributed by vt_m.`

## PHP

 ``

## Javascript

 ``

Output

`4290`

Time Complexity: O(n)

Auxiliary Space: O(1)
Method 2: In this case we use formula to add sum of series.

``` Given series 1*2*3 + 2*3*4 + 3*4*5 + 4*5*6 + . . . + n*(n+1)*(n+2)
sum of series = (n * (n+1) * (n+2) * (n+3)) / 4

Put n = 10 then
sum = (10 * (10+1) * (10+2) * (10+3)) / 4
= (10 * 11 * 12 * 13) / 4
= 4290```

## C++

 `// Program to find the sum of series` `// 1*2*3 + 2*3*4 + . . . + n*(n+1)*(n+1)` `#include ` `using` `namespace` `std;`   `// Function to calculate sum of series.` `int` `sumOfSeries(``int` `n)` `{` `    ``return` `(n * (n + 1) * (n + 2) * (n + 3)) / 4;` `}`   `// Driver function` `int` `main()` `{` `    ``int` `n = 10;` `    ``cout << sumOfSeries(n);` `    ``return` `0;` `}`

## Java

 `// Program to find the ` `// sum of series` `// 1*2*3 + 2*3*4 +` `// . . . + n*(n+1)*(n+1)` `import` `java.io.*;`   `class` `GFG {` `    `  `    ``// Function to calculate` `    ``// sum of series.` `    ``static` `int` `sumOfSeries(``int` `n)` `    ``{` `        ``return` `(n * (n + ``1``) * ` `            ``(n + ``2``) * (n + ``3``)) / ``4``;` `    ``}`   `    ``// Driver function` `    ``public` `static` `void` `main (String[] args) {` `        ``int` `n = ``10``;` `        ``System.out.println(sumOfSeries(n));` `        `  `    ``}` `}`   `// This code is contributed by Nikita Tiwari.`

## Python3

 `# Python program to find the ` `# sum of series` `# 1*2*3 + 2*3*4 + . . . ` `# + n*(n+1)*(n+1)`   `# Function to calculate sum ` `# of series.` `def` `sumOfSeries(n):` `    ``return` `(n ``*` `(n ``+` `1``) ``*` `(n ``+` `2` `                    ``) ``*` `(n ``+` `3``)) ``/` `4`   `#Driver code` `n ``=` `10` `print``(sumOfSeries(n))`   `# This code is contributed by "Abhishek Sharma 44"`

## C#

 `// Program to find the ` `// sum of series` `// 1*2*3 + 2*3*4 +` `// . . . + n*(n+1)*(n+1)` `using` `System;`   `class` `GFG {` `    `  `    ``// Function to calculate` `    ``// sum of series.` `    ``static` `int` `sumOfSeries(``int` `n)` `    ``{` `        ``return` `(n * (n + 1) * ` `            ``(n + 2) * (n + 3)) / 4;` `    ``}`   `    ``// Driver function` `    ``public` `static` `void` `Main () {` `        ``int` `n = 10;` `        ``Console.WriteLine(sumOfSeries(n));` `        `  `    ``}` `}`   `// This code is contributed by vt_m.`

## PHP

 ``

## Javascript

 ``

Output

`4290`

Time Complexity: O(1)

Auxiliary Space: O(1)

How does this formula work?

```We can prove working of this formula using
mathematical induction.

According to formula, sum of (k -1) terms is
((k - 1) * (k) * (k + 1) * (k + 2)) / 4

Sum of k terms
= sum of k-1 terms + value of k-th term
= ((k - 1) * (k) * (k + 1) * (k + 2)) / 4 +
k * (k + 1) * (k + 2)
Taking common term (k + 1) * (k + 2) out.
= (k + 1)*(k + 2) [k*(k-1)/4 + k]
= (k + 1)*(k + 2) * k * (k + 3)/4
= k * (k + 1) * (k + 2) * (k + 3)/4```

Avoiding the overflow:
In the above method, sometimes due to the large value of n, the value of (n * (n + 1) * (n + 2) * (n + 3)) would overflow. We can avoid this overflow to some extent using the fact that n*(n+1) must be divisible by 2 and (n+2)*(n+3) is also divisible by 2.

Below is the code for the above approach.

## C++

 `// Program to find the sum of series` `// 1*2*3 + 2*3*4 + . . . + n*(n+1)*(n+1)` `#include ` `using` `namespace` `std;`   `// Function to calculate sum of series.` `int` `sumOfSeries(``int` `n)` `{` `    ``return` `((n * (n + 1) /2)* ((n + 2) * (n + 3)/2));` `}`   `// Driver function` `int` `main()` `{` `    ``int` `n = 10;` `    ``cout << sumOfSeries(n);` `    ``return` `0;` `}`

## Java

 `// Java Program to find the sum of series` `// 1*2*3 + 2*3*4 + . . . + n*(n+1)*(n+1)` `import` `java.io.*;`   `class` `GFG {` `    `  `    ``// Function to calculate sum of series.` `    ``static` `int` `sumOfSeries(``int` `n)` `    ``{` `        ``return` `((n * (n + ``1``) /``2``)* ((n + ``2``) * (n + ``3``)/``2``));` `    ``}`   `    ``// Driver function` `    ``public` `static` `void` `main (String[] args) {` `        ``int` `n = ``10``;` `        ``System.out.println(sumOfSeries(n));` `        `  `    ``}` `}`   `// This code is contributed by Aman Kumar.`

## Python3

 `# Python program to find the sum of series` `# 1*2*3 + 2*3*4 + . . . + n*(n+1)*(n+1)`   `# Function to calculate sum of series` `def` `sumOfSeries(n):` `    ``return` `((n ``*` `(n ``+` `1``) ``/``/` `2``) ``*` `((n ``+` `2``) ``*` `(n ``+` `3``) ``/``/` `2``))`   `# Driver function` `if` `__name__ ``=``=` `'__main__'``:` `    ``n ``=` `10` `    ``print``(sumOfSeries(n))`

## C#

 `// C# program to find the sum of series` `// 1*2*3 + 2*3*4 + . . . + n*(n+1)*(n+1)`   `// Function to calculate sum of series` `using` `System;`   `class` `Program {` `    ``static` `int` `SumOfSeries(``int` `n) {` `        ``return` `((n * (n + 1) / 2) * ((n + 2) * (n + 3) / 2));` `    ``}`   `    ``// Driver function` `    ``static` `void` `Main(``string``[] args) {` `        ``int` `n = 10;` `        ``Console.WriteLine(SumOfSeries(n));` `    ``}` `}`   `// This code is contributed by shiv1o43g`

## Javascript

 `// JavaScript program to find the sum of series` `// 1*2*3 + 2*3*4 + . . . + n*(n+1)*(n+1)`   `// Function to calculate sum of series` `function` `sumOfSeries(n) {` `    ``return` `((n * (n + 1) / 2) * ((n + 2) * (n + 3) / 2));` `}`   `// Driver function` `let n = 10;` `console.log(sumOfSeries(n));`

Output

`4290`

Time complexity: O(1)

Auxiliary Space: O(1)

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