1. Abridged Problem Statement  
Given an integer K (1 ≤ K ≤ 40), compute the sum S = F₁ + F₂ + … + F_K of the first K Fibonacci numbers, where F₁ = 1, F₂ = 1, and Fₙ = Fₙ₋₁ + Fₙ₋₂ for n > 2.

2. Detailed Editorial  
Definition and Constraints  
- Fibonacci sequence: F₁ = 1, F₂ = 1, Fₙ = Fₙ₋₁ + Fₙ₋₂.  
- Input K up to 40, so any O(K) approach is instantaneous.

Two straightforward methods:

Method A: Iteration and Summation  
1. Initialize two variables a = F₁ = 1, b = F₂ = 1.  
2. Initialize sum S = a + b (if K ≥ 2; handle K = 1 separately).  
3. For i from 3 to K, compute c = a + b, add c to S, then shift (a ← b, b ← c).  
4. Output S.

Method B: Closed-form identity  
It is known that  
  S = F₁ + F₂ + … + F_K = F_{K+2} − 1.  
Hence one can compute F_{K+2} by iteration (or fast doubling) and subtract 1.  
Since K ≤ 40, simple iteration is easiest.

Implementation Details  
- Read integer K.  
- If K = 1, answer is 1.  
- Otherwise run an O(K) loop to build Fibonacci numbers up to F_K (and accumulate sum).  
- Print result as a 64-bit integer (sum can be as large as about F₄₂ ~ 2.6×10⁸).

3. Provided C++ Solution with Detailed Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

// We store Fibonacci numbers up to 40 in array f,
// and keep a running total 'sum'.
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n;
    cin >> n;               // Read input K (denoted n here)

    // f[i] will hold the i-th Fibonacci number.
    // We need up to f[n], but array size is set to handle up to 40.
    static long long f[1 << 10];
    long long sum = 0;

    // Base cases:
    // f[0] is unused; define f[1] = 1, f[2] = 1
    if (n >= 1) f[1] = 1;
    if (n >= 2) f[2] = 1;

    // Sum the base cases if they exist
    if (n >= 1) sum += f[1];
    if (n >= 2) sum += f[2];

    // Build Fibonacci numbers from 3 to n, and add each to sum
    for (int i = 3; i <= n; i++) {
        f[i] = f[i - 1] + f[i - 2];  // Fibonacci recurrence
        sum += f[i];                 // Accumulate into sum
    }

    // Output the final sum
    cout << sum << "\n";
    return 0;
}
```

4. Python Solution with Detailed Comments  
```python
def main():
    import sys
    data = sys.stdin.read().strip().split()
    if not data:
        return
    k = int(data[0])  # Number of Fibonacci terms to sum

    # Handle small k directly
    if k == 1:
        print(1)
        return

    # Initialize first two Fibonacci numbers
    a, b = 1, 1
    total = a + b  # Sum of F1 and F2

    # Generate F3..Fk and accumulate their sum
    for _ in range(3, k + 1):
        c = a + b    # Next Fibonacci number
        total += c   # Add to running total
        a, b = b, c  # Shift for next iteration

    print(total)

if __name__ == "__main__":
    main()
```

5. Compressed Editorial  
Compute Fibonacci up to K in O(K), keep a running sum. Base cases F₁ = F₂ = 1, then loop i=3…K: Fᵢ = Fᵢ₋₁ + Fᵢ₋₂, sum+=Fᵢ. Output sum. Alternatively use S = F_{K+2} − 1.