Find nth term of a given recurrence relation in C++

Concept

Assume bn be a sequence of numbers, which is denoted by the recurrence relation b₁=1 and b_n+1/b_n=2ⁿ. Our task is to determine the value of log₂(b_n) for a given n.

Input

6

Output

15

Explanation

log₂(b_n) = (n * (n - 1)) / 2 = (6*(6-1))/2 = 15

Input

200

Output

19900

Method

b_n+1/b_n = 2ⁿ

b_n/b_n-1 = 2^n-1

.

.

.

b₂/b₁ = 2¹, We multiply all of above in order to attain

(b_n+1/b_n).(b_n/_n-1)……(b₂/b₁) = 2^{n + (n-1)+……….+1}

So, b_n+1/b₁ = 2^n(n+1)/2

Because we know, 1 + 2 + 3 + ………. + (n-1) + n = n(n+1)/2

So, b_n+1 = 2^n(n+1)/2 . b₁; Assume the initial value b₁ = 1

So, b_n+1 = 2sup>n(n+1)/2

Now substituting (n+1) for n, we get,

b_n = 2^n(n-1)/2

Taking log both sides, we get,

log₂(b_n) = n(n-1)/2

Example

 Live Demo

// C++ program to find nth term of
// a given recurrence relation
#include <bits/stdc++.h>
using namespace std;
// Shows function to return required value
int sum(int n1){
   // Now get the answer
   int ans1 = (n1 * (n1 - 1)) / 2;
   //Now return the answer
   return ans1;
}
// Driver program
int main(){
   // Get the value of n
   // int n = 6;
   int n = 200;
   // Uses function call to print result
   cout << sum(n);
   return 0;
}

Output

19900