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 sumint sum(int n){ if (n == 1) { return 2; } else { return (n * (n + 1) + sum(n - 1)); }}// Driver codeint main(){ int n = 2; cout << sum(n);} |
Java
// Java implementation of the approachclass 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 sumdef sum(n): if (n == 1): return 2; else: return (n * (n + 1) + sum(n - 1));# Driver coden = 2;print(sum(n));# This code is contributed by mits |
C#
// Csharp implementation of the approachusing 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 sumfunction sum($n){ if ($n == 1) { return 2; } else { return ($n * ($n + 1) + sum($n - 1)); }}// Driver code$n = 2;echo sum($n);// This code is contributed by mits?> |
Javascript
<script>// JavaScript implementation of the approach// Function to return sumfunction sum(n){ if (n == 1) { return 2; } else { return (n * (n + 1) + sum(n - 1)); }}// Driver coden = 2;document.write(sum(n));// This code is contributed by rutvik_56.</script> |
Output:
8
Time Complexity: O(n)
Auxiliary Space: O(n), since n extra space has been taken.
Efficient Solution We can solve this problem using direct formula.
Sum can be written as below
?(n * (n+1))
?(n*n + n)
= ?(n*n) + ?(n)
We can apply the formulas for sum squares of natural number and sum of natural numbers.
= n(n+1)(2n+1)/6 + n*(n+1)/2
= n * (n + 1) * (n + 2) / 3
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return sumint sum(int n){ return n * (n + 1) * (n + 2) / 3;}// Driver codeint main(){ int n = 2; cout << sum(n);} |
Java
// Java implementation of the approachclass GFG{ // Function to return sumstatic int sum(int n){ return n * (n + 1) * (n + 2) / 3;}// Driver codepublic static void main(String[] args){ int n = 2; System.out.println(sum(n));}}// This code is contributed by Code_Mech |
Python3
# Python3 implementation of the approach. # Function to return sum def Sum(n): return n * (n + 1) * (n + 2) // 3# Driver code if __name__ == "__main__": n = 2; print(Sum(n)) # This code is contributed # by Rituraj Jain |
C#
// C# implementation of the approachusing System; class GFG{ // Function to return sumstatic int sum(int n){ return n * (n + 1) * (n + 2) / 3;} // Driver codepublic static void Main(String[] args){ int n = 2; Console.WriteLine(sum(n));}}// This code contributed by Rajput-Ji |
PHP
<?php// PHP implementation of the approach // Function to return sum function sum($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 sumfunction sum(n){ return n * (n + 1) * (n + 2) / 3;}// Driver codevar n = 2;document.write(sum(n));// This code is contributed by noob2000.</script> |
Output:
8
Time Complexity: O(1)
Auxiliary Space: O(1)
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