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

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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