If $(1+x+x^2)^n = a_0 + a_1x + a_2x^2 + ... + a_{2n}x^{2n}$, then the value of $a_1 - a_3 + a_5 - a_7 + ... $ is equal to :
Step-by-Step Solution
Key Concept: Using complex number substitution with $x = i$ efficiently extracts coefficients of odd powers with alternating signs based on powers of $i$ cycling through period 4.
To find $a_1 - a_3 + a_5 - a_7 + ...$, we use the substitution method. Let $P(x) = (1+x+x^2)^n = \sum a_i x^i$. We evaluate $P(i) - P(-i)$ where $i = \sqrt{-1}$ to extract odd-indexed coefficients with alternating signs. We have $P(i) = (1+i+i^2)^n = (1+i-1)^n = i^n$ and $P(-i) = (1-i-1)^n = (-i)^n$. Therefore, $\frac{P(i) - P(-i)}{2i} = a_1 - a_3 + a_5 - ... = \frac{i^n - (-i)^n}{2i}$. This value depends on $n \bmod 4$: when $n \equiv 0 \pmod{4}$, we get $1$; when $n \equiv 1 \pmod{4}$, we get $-1$; when $n \equiv 2 \pmod{4}$, we get $0$; when $n \equiv 3 \pmod{4}$, we get $-1$.
Correct Answer: 1,2,3,4