C++ Program To Find Sum of Fibonacci Numbers at Even Indexes Upto N Terms

Given a positive integer N, the task is to find the value of F₂ + F₄ + F₆ +………+ F_2n upto N terms where F_i denotes the i-th Fibonacci number.
The Fibonacci numbers are the numbers in the following integer sequence.

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ……

Examples:

Input: n = 5 
Output: 88 
N = 5, So the fibonacci series will be generated from 0th term upto 10th term: 
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 
Sum of elements at even indexes = 0 + 1 + 3 + 8 + 21 + 55
Input: n = 8 
Output: 1596 
0 + 1 + 3 + 8 + 21 + 55 + 144 + 377 + 987 = 1596.

Method-1: This method includes solving the problem directly by finding all Fibonacci numbers till 2n and adding up the only the even indices. But this will require O(n) time complexity.
Below is the implementation of the above approach:

C++

// C++ Program to find sum of
// even-indiced Fibonacci numbers
#include <bits/stdc++.h>
using namespace std;

// Computes value of first
// fibonacci numbers and
// stores the even-indexed sum
int calculateEvenSum(int n)
{
    if (n <= 0)
        return 0;

    int fibo[2 * n + 1];
    fibo[0] = 0, fibo[1] = 1;

    // Initialize result
    int sum = 0;

    // Add remaining terms
    for (int i = 2; i <= 2 * n; i++)
    {
        fibo[i] = fibo[i - 1] + fibo[i - 2];

        // For even indices
        if (i % 2 == 0)
            sum += fibo[i];
    }

    // Return the alternating sum
    return sum;
}

// Driver code
int main()
{

    // Get n
    int n = 8;

    // Find the even-indiced sum
    cout << "Even indexed Fibonacci Sum upto " <<
             n << " terms: " << calculateEvenSum(n) <<
             endl;

    return 0;
}

Output:

Even indexed Fibonacci Sum upto 8 terms: 1596

Time complexity: O(N).
Auxiliary space: O(N).

Method-2:

It can be clearly seen that the required sum can be obtained thus: 
2 ( F₂ + F₄ + F₆ +………+ F_2n ) = (F₁ + F₂ + F₃ + F₄ +………+ F_2n) – (F₁ – F₂ + F₃ – F₄ +………+ F_2n)
Now the first term can be obtained if we put 2n instead of n in the formula given here.
Thus F₁ + F₂ + F₃ + F₄ +………+ F_2n = F_2n+2 – 1.
The second term can also be found if we put 2n instead of n in the formula given here
Thus, F₁ – F₂ + F₃ – F₄ +………- F_2n = 1 + (-1)²ⁿ⁺¹F_2n-1 = 1 – F_2n-1.
So, 2 ( F₂ + F₄ + F₆ +………+ F_2n) 
= F_2n+2 – 1 – 1 + F_2n-1
= F_2n+2 + F_2n-1 – 2 
= F_2n + F_2n+1 + F_2n+1 – F_2n – 2 
= 2 ( F_2n+1 -1) 
Hence, ( F₂ + F₄ + F₆ +………+ F_2n) = F_2n+1 -1 .

So in order to find the required sum, the task is to find only F_2n+1 which requires O(log n) time.( Refer to method 5 or method 6 in this article.
Below is the implementation of the above approach:

C++

// C++ Program to find even indexed
// Fibonacci Sum in O(Log n) time.
#include <bits/stdc++.h>
using namespace std;

const int MAX = 1000;

// Create an array for memoization
int f[MAX] = { 0 };

// Returns n'th Fibonacci number
// using table f[]
int fib(int n)
{
    // Base cases
    if (n == 0)
        return 0;
    if (n == 1 || n == 2)
        return (f[n] = 1);

    // If fib(n) is already computed
    if (f[n])
        return f[n];

    int k = ((n & 1) ? (n + 1) / 2 :
              n / 2);

    // Applying above formula [Note value
    // n&1 is 1 if n is odd, else 0].
    f[n] = (n & 1) ? (fib(k) * fib(k) +
                      fib(k - 1) * fib(k - 1)) :
                     (2 * fib(k - 1) + fib(k)) *
                      fib(k);

    return f[n];
}

// Computes value of even-indexed
// Fibonacci Sum
int calculateEvenSum(int n)
{
    return (fib(2 * n + 1) - 1);
}

// Driver code
int main()
{
    // Get n
    int n = 8;

    // Find the alternating sum
    cout << "Even indexed Fibonacci Sum upto " <<
             n << " terms: " << calculateEvenSum(n) <<
             endl;

    return 0;
}

Output:

Even indexed Fibonacci Sum upto 8 terms: 1596

Time complexity: O(log(N)).
Auxiliary Space: O(N).