[Solved] Calculating without running the program [closed]

Question

The x += x doubles x every time the innermost statement is executed.
The inner loop is executed 1 + 2 + 3 + 4 = 4 * (4 + 1) / 2 = 10 times. Thus, the result must be 2^10 = 1024.

For general n, it should be something like 2^((n + 1) * n / 2).

The fact that 1 + 2 + ... + n = (n + 1) * n / 2 is sometimes called “Gaussian sum formula”, you should remember it next time you see two nested loops where the range of the inner index depends on the outer index.

Accepted Answer

The x += x doubles x every time the innermost statement is executed.
The inner loop is executed 1 + 2 + 3 + 4 = 4 * (4 + 1) / 2 = 10 times. Thus, the result must be 2^10 = 1024.

For general n, it should be something like 2^((n + 1) * n / 2).

The fact that 1 + 2 + ... + n = (n + 1) * n / 2 is sometimes called “Gaussian sum formula”, you should remember it next time you see two nested loops where the range of the inner index depends on the outer index.