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Program to find sum of series 1*2*3 + 2*3*4+ 3*4*5 + . . . + n*(n+1)*(n+2)
• Difficulty Level : Easy
• Last Updated : 24 Apr, 2018

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
```

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

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)); ` `    ``} ` `}  ` ` `  `// This article is contributed by Gitanjali. `

## 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

 ` `

Output:

```4290
```

Time Complexity: O(n)

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

 ` `

Output:

```4290
```

Time Complexity: 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```

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