Given $(1 + x + 2x^2)^{11} = 1 + a_1x + a_2x^2 + \ldots + a_{12}x^{12}$, find $a_1 + a_2 + \ldots + a_{12}$.
Step-by-Step Solution
Key Concept: Substitute specific values of $x$ (typically 1 or -1) into the polynomial expansion to find sums of coefficients
On putting $x = 1$ in the equation $(1 + x + 2x^2)^{11} = 1 + a_1x + a_2x^2 + \ldots + a_{12}x^{12}$, we get $(1 + 1 + 2)^{11} = 1 + a_1 + a_2 + \ldots + a_{12}$, which gives $4^{11} = 1 + a_1 + a_2 + \ldots + a_{12}$. Therefore, $a_1 + a_2 + \ldots + a_{12} = 4^{11} - 1 = 3^{11}$. Alternatively, on putting $x = -1$, we get $0^{11} = 1 + a_1(-1) + a_2(-1)^2 + \ldots + a_{12}(-1)^{12}$.
Correct Answer: 3^{11} - 1