import java.util.Scanner;

public class Solution {

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

    while (t-- > 0) {
        long n = in.nextLong();
        long a = 1, b = 2, sum = 0;


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


            long temp = a + b;
            a = b;
            b = temp;
        }


        System.out.println(sum);
    }

    in.close();
} 

}

def sum_fib(n): a,b= 1,2 even_sum = 0 while b<=n: if b%2==0: even_sum +=b a,b=b,a+b return even_sum

if name == 'main': t = int(input().strip())

for t_itr in range(t):
    n = int(input().strip())
    res = sum_fib(n)
    print(res)

#!/bin/python3

import sys

t = int(input().strip())
for a0 in range(t):
    n = int(input().strip())
    def fibonaci(n):
        if n == 1:
            return 1
        if n <= 0:
            return 0
        a, b = 0, 1  # Defining the 1st and 2nd position values of the Fibonacci series
        ans = 0
        while True:  # Creating a loop to update the values of a and b so that we can check if the next value is even or not
            if a % 2 == 0:  # Considering a to always be the first value and checking if it is even or not
                ans += a  # If the value stored in a is even, add it to the result
            a, b = b, a + b  # Swap the values: set a to b and b to the sum of a and b as per the Fibonacci series (e.g., 0 1 1 2 3 5 8 -> a=0 b=1 -> step 2 a=1 b=1 -> step 3 a=1 b=2 and so on)
            if a > n:  # Check if the value of a exceeds the given n, and if so, break the loop
                break
        print(ans)  # Output the final answer
    fibonaci(n)

For the sake of discussion and explanation of the question, I am providing detailed comments within the code. Please refer to these comments for a clear understanding of the logic and functionality of the code.

**//1,1,2,3,5,8,13,21,34,55,89,144 //using the knowledge that after every two terms comes an even term //Sum of two odds is an even(two oods then an even) ** int main(){ int t; cin >> t; for(int a0 = 0; a0 < t; a0++){ long n; cin >> n; long a1=1; long a2=1;

    long sum=a1+a2;
    while(a1+2*a2<n){
        long x=2*a2+a1;
        a2=x+a1+a2;
        a1=x;
        sum+=a1+a2;

    }
            //if the current term was greater than n, subtract their sum from the overall sum
    if(a1+a2>n){
        sum-=(a1+a2);
    }
    cout<<sum<<endl;
}
return 0;

}

KOTLIN Code

fun main(args: Array<String>) {
    val t = readLine()!!.trim().toInt()

    for (tItr in 1..t) {
        val n = readLine()!!.trim().toLong()
        createFibonacciSum(n)
    } 
}

private fun createFibonacciSum(n: Long){
    var left = 0L
    var middle = 1L
    var sum = 0L
    var i = 0
    while(i< n && (middle + left)< n){
        var right = middle + left
        left = middle
        middle = right
        if(right % 2L == 0L) sum += right
        i++ 
    }
    println(sum)
}