# Find nth term of a given recurrence relation

Let an be a sequence of numbers, which is defined by the recurrence relation a1=1 and an+1/an=2n. The task is to find the value of log2(an) for a given n.

Examples:

Input: 5
Output: 10
Explanation:
log2(an) = (n * (n - 1)) / 2
= (5*(5-1))/2
= 10

Input: 100
Output: 4950


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.      , We multiply all of the above in order to reach  Since . Then .
Substituting n+1 for n: So, Below is the implementation of above approach.

## C++

 // C++ program to find nth term of   // a given recurrence relation     #include  using namespace std;     // function to return required value  int sum(int n)  {         // Get the answer      int ans = (n * (n - 1)) / 2;         // Return the answer      return ans;  }     // Driver program  int main()  {         // Get the value of n      int n = 5;         // function call to print result      cout << sum(n);         return 0;  }

## Java

 // Java program to find nth term   // of a given recurrence relation  import java.util.*;     class solution  {  static int sum(int n)  {      // Get the answer      int ans = (n * (n - 1)) / 2;         // Return the answer      return ans;  }     // Driver code  public static void main(String arr[])  {         // Get the value of n      int n = 5;         // function call to print result      System.out.println(sum(n));  }  }  //This code is contributed byte  //Surendra_Gangwar

## Python3

 # Python3 program to find nth   # term of a given recurrence   # relation     # function to return   # required value  def sum(n):         # Get the answer      ans = (n * (n - 1)) / 2;             # Return the answer      return ans     # Driver Code     # Get the value of n  n = 5    # function call to prresult  print(int(sum(n)))     # This code is contributed by Raj

## C#

 // C# program to find nth term   // of a given recurrence relation  using System;     class GFG  {  static int sum(int n)  {      // Get the answer      int ans = (n * (n - 1)) / 2;         // Return the answer      return ans;  }     // Driver code  public static void Main()  {         // Get the value of n      int n = 5;         // function call to print result      Console.WriteLine(sum(n));  }  }     // This code is contributed byte  // inder_verma

## PHP

 

Output:

10


Time Complexity: O(1)

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