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Java Program to Find Sum of Fibonacci Series Numbers of First N Even Indexes
• Last Updated : 17 Mar, 2021

For a given positive integer N, the purpose is to find the value of F2 + F4 + F6 +………+ F2n till N number. Where Fi indicates the i’th Fibonacci number.

The Fibonacci Series are the numbers in the below-given integer sequence.

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

Examples:

```Input: n = 4
Output: 33
N = 4, So here the fibonacci series will be produced from 0th term till 8th term:
0, 1, 1, 2, 3, 5, 8, 13, 21
Sum of numbers at even indexes = 0 + 1 + 3 + 8 + 21 = 33.

Input: n = 7
Output: 609
0 + 1 + 3 + 8 + 21 + 55 + 144 + 377 = 609.```

Approach 1:

Find all Fibonacci numbers till 2n and adding up only the even indices.

## Java

 `// Java Program to find even sum of``// fibonacci Series Till number N``import` `java.io.*;`` ` `class` `geeksforgeeks {`` ` `    ``// Computing the value of first fibonacci series``    ``// and storing the sum of even indexed numbers``    ``static` `int` `Fib_Even_Sum(``int` `N)``    ``{``        ``if` `(N <= ``0``)``            ``return` `0``;`` ` `        ``int` `fib[] = ``new` `int``[``2` `* N + ``1``];``        ``fib[``0``] = ``0``;``        ``fib[``1``] = ``1``;`` ` `        ``// Initializing the sum``        ``int` `s = ``0``;`` ` `        ``// Adding remaining numbers``        ``for` `(``int` `j = ``2``; j <= ``2` `* N; j++) {``            ``fib[j] = fib[j - ``1``] + fib[j - ``2``];`` ` `            ``// Only considering even indexes``            ``if` `(j % ``2` `== ``0``)``                ``s += fib[j];``        ``}`` ` `        ``return` `s;``    ``}`` ` `    ``// The Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `N = ``11``;`` ` `        ``// Prints the sum of even-indexed numbers``        ``System.out.println(``            ``"Even sum of fibonacci series till number "` `+ N``            ``+ ``" is: "` `+ +Fib_Even_Sum(N));``    ``}``}`
Output
```Even sum of fibonacci series till number 11 is: 28656
```

Time Complexity: O(n)

Approach 2:

It can be clearly seen that the required sum can be obtained thus:
2 ( F2 + F4 + F6 +………+ F2n ) = (F1 + F2 + F3 + F4 +………+ F2n) – (F1 – F2 + F3 – F4 +………+ F2n)

Now the first term can be obtained if we put 2n instead of n in the formula given here.

Thus F1 + F2 + F3 + F4 +………+ F2n = F2n+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 +………- F2n = 1 + (-1)2n+1F2n-1 = 1 – F2n-1.

So, 2 ( F2 + F4 + F6 +………+ F2n)
= F2n+2 – 1 – 1 + F2n-1
= F2n+2 + F2n-1 – 2
= F2n + F2n+1 + F2n+1 – F2n – 2
= 2 ( F2n+1 -1)
Hence, ( F2 + F4 + F6 +………+ F2n) = F2n+1 -1 .

The task is to find only F2n+1 -1.

Below is the implementation of the above approach:

## Java

 `// Java Program to find even indexed``// Fibonacci Sum in O(Log n) time.`` ` `class` `GFG {`` ` `    ``static` `int` `MAX = ``1000``;`` ` `    ``// Create an array for memoization``    ``static` `int` `f[] = ``new` `int``[MAX];`` ` `    ``// Returns n'th Fibonacci number``    ``// using table f[]``    ``static` `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] == ``1``) {``            ``return` `f[n];``        ``}`` ` `        ``int` `k = (n % ``2` `== ``1``) ? (n + ``1``) / ``2` `: n / ``2``;`` ` `        ``// Applying above formula [Note value n&1 is 1``        ``// if n is odd, else 0].``        ``f[n] = (n % ``2` `== ``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``    ``static` `int` `calculateEvenSum(``int` `n)``    ``{``        ``return` `(fib(``2` `* n + ``1``) - ``1``);``    ``}`` ` `    ``// Driver program to test above function``    ``public` `static` `void` `main(String[] args)``    ``{``        ``// Get n``        ``int` `n = ``11``;`` ` `        ``// Find the alternating sum``        ``System.out.println(``            ``"Even indexed Fibonacci Sum upto "` `+ n``            ``+ ``" terms: "` `+ calculateEvenSum(n));``    ``}``}`
Output
```Even indexed Fibonacci Sum upto 8 terms: 1596
```

Time Complexity: O(log n)

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