Binomial Theorem
Sum of Coefficients
Grade 11
Question:
<p>If \((1 - x + x^2)^n = a_0 + a_1 x + a_2 x^2 + \cdots + a_{2n} x^{2n}\), then \(a_0 + a_2 + a_4 + \cdots + a_{2n}\) is equal to</p>
<p>(a) \(\frac{1 + 3^n}{2}\)</p>
<p>(b) \(\frac{1 + 3^n}{2}\)</p>
<p>(c) \(\frac{1 - 3^n}{2}\)</p>
<p>(d) \(3 + \frac{n}{2}\)</p>
Step-by-Step Solution
Key Concept: Substitute $x = 1$ and $x = -1$ into the expansion, then average to isolate the sum of even-indexed coefficients.
<p><strong>Solution:</strong> Let $f(x) = (1 - x + x^2)^n = a_0 + a_1 x + a_2 x^2 + \cdots + a_{2n} x^{2n}$</p><p>To extract even coefficients: $f(1) = (1 - 1 + 1)^n = 1$ and $f(-1) = (1 + 1 + 1)^n = 3^n$</p><p>Then $a_0 + a_2 + a_4 + \cdots + a_{2n} = \frac{f(1) + f(-1)}{2} = \frac{1 + 3^n}{2}$</p>
Correct Answer: A