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Sum of the series (1*2) + (2*3) + (3*4) + …… upto n terms
  • Last Updated : 18 Mar, 2021

Given a value n, the task is to find sum of the series (1*2) + (2*3) + (3*4) + ……+ n terms
Examples: 
 

Input: n = 2
Output: 8
Explanation:
(1*2) + (2*3)
= 2 + 6
= 8

Input: n = 3
Output: 20
Explanation:
(1*2) + (2*3) + (2*4)
= 2 + 6 + 12
= 20

 

Simple Solution One by one add elements recursively.
Below is the implementation 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return sum
int sum(int n)
{
    if (n == 1) {
        return 2;
    }
    else {
        return (n * (n + 1) + sum(n - 1));
    }
}
 
// Driver code
int main()
{
 
    int n = 2;
    cout << sum(n);
}

Java




// Java implementation of the approach
 
class Solution {
 
    // Function to return a the required result
    static int sum(int n)
    {
        if (n == 1) {
            return 2;
        }
        else {
            return (n * (n + 1) + sum(n - 1));
        }
    }
    // Driver code
    public static void main(String args[])
    {
        int n = 2;
        System.out.println(sum(n));
    }
}

Python3




# Python3 implementation of the approach
 
# Function to return sum
def sum(n):
 
    if (n == 1):
        return 2;
    else:
        return (n * (n + 1) + sum(n - 1));
 
# Driver code
 
n = 2;
print(sum(n));
 
# This code is contributed by mits

C#




// Csharp implementation of the approach
 
using System;
 
class Solution {
 
    // Function to return a the required result
    static int sum(int n)
    {
        if (n == 1) {
            return 2;
        }
        else {
            return (n * (n + 1) + sum(n - 1));
        }
    }
    // Driver code
    public static void Main()
    {
        int n = 2;
        Console.WrieLine(sum(n));
    }
}

PHP




<?php
// PHP implementation of the approach
 
// Function to return sum
function sum($n)
{
    if ($n == 1)
    {
        return 2;
    }
    else
    {
        return ($n * ($n + 1) +
                  sum($n - 1));
    }
}
 
// Driver code
$n = 2;
echo sum($n);
 
// This code is contributed by mits
?>

Javascript




<script>
 
// JavaScript implementation of the approach
 
// Function to return sum
function sum(n)
{
    if (n == 1) {
        return 2;
    }
    else {
        return (n * (n + 1) + sum(n - 1));
    }
}
 
// Driver code
n = 2;
document.write(sum(n));
 
// This code is contributed by rutvik_56.
 
</script>
Output: 
8

 

Time Complexity: O(n)
Efficient Solution We can solve this problem using direct formula. 
Sum can be written as below 
∑(n * (n+1)) 
∑(n*n + n) 
= ∑(n*n) + ∑(n)
We can apply the formulas for sum squares of natural number and sum of natural numbers.
= n(n+1)(2n+1)/6 + n*(n+1)/2 
= n * (n + 1) * (n + 2) / 3
 



C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return sum
int sum(int n)
{
    return n * (n + 1) * (n + 2) / 3;
}
 
// Driver code
int main()
{
    int n = 2;
    cout << sum(n);
}

Java




// Java implementation of the approach
class GFG
{
     
// Function to return sum
static int sum(int n)
{
    return n * (n + 1) * (n + 2) / 3;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 2;
    System.out.println(sum(n));
}
}
 
// This code is contributed by Code_Mech

Python3




# Python3 implementation of the approach.
 
# Function to return sum
def Sum(n):
 
    return n * (n + 1) * (n + 2) // 3
 
# Driver code
if __name__ == "__main__":
 
    n = 2;
    print(Sum(n))
 
# This code is contributed
# by Rituraj Jain

C#




// C# implementation of the approach
using System;
     
class GFG
{
      
// Function to return sum
static int sum(int n)
{
    return n * (n + 1) * (n + 2) / 3;
}
  
// Driver code
public static void Main(String[] args)
{
    int n = 2;
    Console.WriteLine(sum(n));
}
}
// This code contributed by Rajput-Ji

PHP




<?php
// PHP implementation of the approach
 
// Function to return sum
function sum($n)
{
    return $n * ($n + 1) * ($n + 2) / 3;
}
 
// Driver code
$n = 2;
echo sum($n);
 
// This code is contributed
// by 29AjayKumar
?>

Javascript




<script>
// Javascript implementation of the approach
 
// Function to return sum
function sum(n)
{
    return n * (n + 1) * (n + 2) / 3;
}
 
// Driver code
var n = 2;
document.write(sum(n));
 
// This code is contributed by noob2000.
</script>
Output: 
8

 

Time Complexity: O(1)
 

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