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.
- Accumulate F₃..F_K in an O(K) loop, then add F₁ and F₂ (each equal to 1) whenever the index is reached.
- Print result as a 64-bit integer (sum can be as large as about F₄₂ ~ 2.6×10⁸).

3. C++ Solution
```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n;

void read() { cin >> n; }

void solve() {
    // Sum of the first K Fibonacci numbers. We accumulate F3..FK iteratively
    // into sum and add F1 = F2 = 1 separately when K reaches those indices.
    // K < 41 so the result fits comfortably in a 64-bit integer.

    int64_t sum = 0;
    int64_t f1 = 1, f2 = 1;
    for(int i = 3; i <= n; i++) {
        int64_t fi = f1 + f2;
        sum += fi;
        f1 = f2;
        f2 = fi;
    }

    if(n >= 1) {
        sum += 1;
    }

    if(n >= 2) {
        sum += 1;
    }

    cout << sum << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    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.
