every even fibinoci number follows a pattern {2,8,34,144} if you look at 8 it is 2*4+0, 34 it is 8*4+2 , 144 it is 34*4+8

n = int(input().strip()) prev=0 i=2 s=0 tenp=0 while i<=n: s+=i temp=i i=i*4+prev prev=temp print(s)

public class Solution {

        public static void main(String[] args) {
                Scanner in = new Scanner(System.in);
                int t = in.nextInt();

                for(int a0 = 0; a0 < t; a0++){
                        long n = in.nextLong();
                        long sum = 0;
                        long x1=1, x2=2;

                        while(x2 <= n){
                                if(x2 % 2 == 0){
                                        sum += x2;
                                }

                                long next = x1 + x2;
                                x1 = x2;
                                x2 = next;   
                        }
                        System.out.println(sum);
                }
        }
}

couple hints: 1. don't compute all fibonacci numbers, the even ones are sufficient 2. store a fine selection of even fibonacci numbers and their pre sums for later reference 3. you can even speed up the search for the matching even fibonacci number by using a matching exponential factor and some log functions

import java.io.; import java.util.; import java.text.; import java.math.; import java.util.regex.*;

public class Solution {

public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int t = in.nextInt();
   for(int a0 = 0; a0 < t; a0++){
        long n = in.nextLong();  // Read the limit N for this test case
        long sum = 0;

        // Starting the Fibonacci sequence
        long a = 1, b = 2;

        // Calculate the sum of even Fibonacci numbers not exceeding N
        while (b <= n) {
            if (b % 2 == 0) {
                sum += b;
            }

            // Generate the next Fibonacci number
            long next = a + b;
            a = b;
            b = next;
        }

        // Output the result for this test case
        System.out.println(sum);
    }

    in.close(); 
}

}

Easier than #1