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

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

C++

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// Program to find the sum of series
// 1*2*3 + 2*3*4 + . . . + n*(n+1)*(n+1)
#include <bits/stdc++.h>
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;
}

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Java

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// 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.

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Python3

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# 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"

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C#

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// 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.

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PHP

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<?php
// PHP 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)
{
    $sum = 0;
    for ($i = 1; $i <= $n; $i++)
        $sum = $sum + $i * ($i + 1) * 
                           ($i + 2);
    return $sum;
}
  
// Driver Code
$n = 10;
echo sumOfSeries($n);
  
// This code is contributed by vt_m.
?>

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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++

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// Program to find the sum of series
// 1*2*3 + 2*3*4 + . . . + n*(n+1)*(n+1)
#include <bits/stdc++.h>
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;
}

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Java

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// 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.

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Python3

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# 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"

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C#

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// 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.

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PHP

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<?php
// PHP 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 * ($n + 1) * ($n + 2) * 
                        ($n + 3)) / 4;
}
  
// Driver Code
$n = 10;
echo sumOfSeries($n);
  
// This code is contributed by vt_m.
?>

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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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Improved By : vt_m