<p>If \((1 + x)^5 = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + a_5 x^5\), then the value of \((a_0 - a_2 + a_4)^2 + (a_1 - a_3 + a_5)^2\) is equal to</p>
Step-by-Step Solution
Key Concept: Substitute x = i (imaginary unit) and x = -i into the expansion to isolate alternating sum patterns, then use |z|² = z·z̄ to evaluate the sum of squares of complex expressions.
<p><strong>Step 1:</strong> From binomial theorem: $(1+x)^5 = \sum_{k=0}^{5} \binom{5}{k}x^k$</p><p>So $a_k = \binom{5}{k}$</p><p><strong>Step 2:</strong> Substitute $x = i$:</p><p>$(1+i)^5 = a_0 + a_1 i - a_2 - a_3 i + a_4 + a_5 i$</p><p>$= (a_0 - a_2 + a_4) + i(a_1 - a_3 + a_5)$</p><p><strong>Step 3:</strong> Calculate $(1+i)^5$:</p><p>$(1+i)^2 = 1 + 2i - 1 = 2i$</p><p>$(1+i)^4 = (2i)^2 = -4$</p><p>$(1+i)^5 = -4(1+i) = -4 - 4i$</p><p><strong>Step 4:</strong> Therefore: $a_0 - a_2 + a_4 = -4$ and $a_1 - a_3 + a_5 = -4$</p><p><strong>Step 5:</strong> $(a_0 - a_2 + a_4)^2 + (a_1 - a_3 + a_5)^2 = (-4)^2 + (-4)^2 = 16 + 16 = 32$</p><p>∴ Answer: <strong>32</strong></p>
Correct Answer: A