The sum of all binomial coefficients in the expansion of $(1+x)^n$ is

1. The sum of binomial coefficients is $\binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n}$. 2. To get this sum, substitute $x = 1$ in the expansion $(1+x)^n = \sum_{r=0}^{n}\binom{n}{r}\,x^r$. 3. LHS becomes $(1+1)^n = 2^n$. RHS becomes $\sum \binom{n}{r}$. 4. So $\sum_{r=0}^{n}\binom{n}{r} = 2^n$. 5. Example check: for $n = 3$, $\binom{3}{0} + \binom{3}{1} + \binom{3}{2} + \binom{3}{3} = 1 + 3 + 3 + 1 = 8 = 2^3$. ✓ 6. Option A ($n^2$) is wrong (only matches for $n = 2, 4$). Option C is half. Option D is the number of TERMS, not the sum. _Source: syllabus Mathematics, Ch 7, §7.2.2 (Some special cases — $x = 1$), p. 5._