Answer: Range $-(2^{n-1}-1)$ to $+(2^{n-1}-1)$

Explanation (n-bit 1's complement):

• Positive numbers: $0$ to $2^{n-1}-1$ (leading bit 0)
• Negative numbers obtained by complementing bits: $-(2^{n-1}-1)$ to $-0$
• Two zeros: $+0 = 00...0$ and $-0 = 11...1$ - wasted code, hence $2^n-1$ effective values but $2^n$ encodings.
• e.g., n=8: $-127$ to $+127$; n=4: $-7$ to $+7$.
• Contrast with 2's complement: $-2^{n-1}$ to $2^{n-1}-1$ (single zero, one extra negative value).