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

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++
Java
Python3
C#
PHP
Javascript
// 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 functionint 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)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.
# 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 coden = 10print(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; 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 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.?>
<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++
Java
Python3
C#
PHP
Javascript
// 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 functionint main(){    int n = 10;    cout << sumOfSeries(n);    return 0;}
// 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.
# 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 coden = 10print(sumOfSeries(n)) # This code is contributed by "Abhishek Sharma 44"
// 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 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.?>
<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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