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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 : 25 Mar, 2021

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 <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;
}

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




<?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.
?>

Javascript




<script>
 
// 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) {
        let sum = 0;
 
        for ( let i = 1; i <= n; i++)
            sum = sum + i * (i + 1) * (i + 2);
 
        return sum;
    }
 
    // Driver Code
        let n = 10;
        document.write(sumOfSeries(n));
 
// This code contributed by Princi Singh
 
</script>

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 <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;
}

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




<?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.
?>

Javascript




<script>
// 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 function
    var n = 10;
    document.write(sumOfSeries(n));
 
// This code is contributed by Amit Katiyar
</script>

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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