If $3^x = a_0 + a_1 + a_2 + \ldots + a_{6n}$ where $a_i$ are the terms in the binomial expansion, find $a_0 + a_2 + a_4 + \ldots + a_{6n}$
Step-by-Step Solution
Key Concept: Use substitution of $x = 1$ and $x = -1$ to separate even and odd indexed terms in a binomial expansion.
Putting $x = 1$, we get $3^n = a_0 + a_1 + a_2 + \ldots + a_{6n}$ ... (1). Putting $x = -1$, we get $1 = a_0 - a_1 + a_2 - \ldots + a_{6n}$ ... (2). Adding equations (1) and (2), we get $a_0 + a_2 + a_4 + \ldots + a_{6n} = \frac{3^n + 1}{2} = \frac{(1+\sqrt{3}^n)(1-\sqrt{3}^n)}{2}$ which simplifies. Therefore $a_0 + a_2 + a_4 + \ldots + a_{6n} = 2K$ where $K$ is an integer, making the sum even.
Correct Answer: even