Sum of all binomial coefficients in (1 + x)ⁿ equals:

To find the sum of all binomial coefficients in the expansion of $(1 + x)^n$, we use the Binomial Theorem: $$(1 + x)^n = \binom{n}{0}x^0 + \binom{n}{1}x^1 + \binom{n}{2}x^2 + \dots + \binom{n}{n}x^n$$ This can be written as: $$(1 + x)^n = C_0 + C_1x + C_2x^2 + \dots + C_nx^n$$ To find the sum of the coefficients ($C_0 + C_1 + C_2 + \dots + C_n$), we need to eliminate the $x$ variables. We do this by substituting $x = 1$ into the identity: $$(1 + 1)^n = C_0 + C_1(1) + C_2(1)^2 + \dots + C_n(1)^n$$ Simplifying the left side: $$2^n = C_0 + C_1 + C_2 + \dots + C_n$$ Therefore, the sum of all binomial coefficients is $2^n$.