Fibonacci sequence is defined as a sequence of integers starting with 1 and 1, where each subsequent value is the sum of the preceding two I.e.

f(0) = 1 f(1) = 1 f(n) = f(n-1) + f(n-2) where n>=2

My goal is to calculate the sum of the first 100 even-values Fibonacci numbers.

So far I’ve found this code which works perfectly to calculate the sum of even numbers to 4million , however I’m unable to find edit the code so that it stops at the sum of the 100th value, rather than reaching 4million.

public class Improvement { public static int Fibonacci(int j) { /** * * Recursive took a long time so continued with iterative * * Complexity is n squared.. try to improve to just n * */ int tmp; int a = 2; int b = 1; int total = 0; do { if(isEven(a)) total +=a; tmp = a + b; b = a; a = tmp; } while (a < j); return total; } private static boolean isEven(int a) { return (a & 1) == 0; } public static void main(String[] args) { // Notice there is no more loop here System.out.println(Fibonacci(4_000_000)); } }

Just to show the console from @mr1554 code answer, the first 100 even values are shown and then the sum of all is 4850741640 as can be seen below:

Any help is appreciated, thanks!

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## Answer

You said.

My goal is to calculate the sum of the first 100 even-values Fibonacci numbers.

That number gets very large very quickly. You need to:

- use BigInteger
- use the mod function to determine if even

For this I could have started from `(1,1)`

but it’s only one term so …

BigInteger m = BigInteger.ZERO; BigInteger n = BigInteger.ONE; BigInteger sumOfEven= BigInteger.ZERO; int count = 0; BigInteger t; while( count < 100) { t = n.add(m); // check if even if (t.mod(BigInteger.TWO).equals(BigInteger.ZERO)) { sumOfEven = sumOfEven.add(t); count++; } n = m; m = t; } System.out.println(sumOfEven);

Prints

290905784918002003245752779317049533129517076702883498623284700

If, on the other hand, from your comment.

My aim is to calculate the sum of the first 100 even numbers

Then you can do that like so

sumFirstNeven = (((2N + 2)N)/2 = (N+1)N so (101)100 = 10100 and the complexity is O(1)

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