The expansion of $(1+x)^n$ at $x = -1$ gives the value

1. Substitute $x = -1$ into $(1+x)^n$: LHS = $(1 + (-1))^n = 0^n = 0$ (for $n \geq 1$). 2. RHS: $\sum_{r=0}^{n}\binom{n}{r}\,(-1)^r = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \binom{n}{3} + \ldots$ 3. So $\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \ldots + (-1)^n\,\binom{n}{n} = 0$. 4. Rearranging: sum of even-indexed coefficients EQUALS sum of odd-indexed coefficients (and each equals $2^{n-1}$). 5. Example check: $n = 3$: $\binom{3}{0} - \binom{3}{1} + \binom{3}{2} - \binom{3}{3} = 1 - 3 + 3 - 1 = 0$ ✓. 6. Option A is $x = 1$ case. Option B and D have no relation to $x = -1$. _Source: NCERT Class 11 Mathematics, Ch 7, §7.2.2 (Special cases — $x = -1$), p. 5–6._