<p>The number of terms in the expansion of \((1 + x)^{101}(1 + x^2 - x)^{100}\) in powers of \(x\) is</p>
Step-by-Step Solution
Key Concept: Expand (1+x)^101(1+x^2-x)^100 by first simplifying (1+x^2-x)^100 = [(1-x)(1+x^2)/(1-x)]^100, then track the maximum and minimum powers of x that can appear. The number of distinct integer powers is the key count.
<p><strong>Step 1:</strong> Rewrite the expression. Note that (1+x^2-x)^100 = [(1-x)(1+x)]^100(1+x)^k for strategic factoring. Actually, observe: (1+x^2-x) = (1-x+x^2). We need the range of powers in (1+x)^101(1+x^2-x)^100.</p><p><strong>Step 2:</strong> From (1+x)^101, we get powers from x^0 to x^101. From (1+x^2-x)^100, the minimum power occurs when we maximize -x terms: minimum is -100x. The maximum power is when we take all x^2 terms: maximum is 200x. So from second factor: x^(-100) to x^200.</p><p><strong>Step 3:</strong> Multiplying these together gives overall powers from x^(-100) to x^(101+200) = x^301. However, we must check which integer powers actually appear. The first factor contributes powers 0,1,2,...,101. The second factor contributes powers that are combinations of 100(2k) + 100m where -100 ≤ m ≤ 100 and 0 ≤ k ≤ 100. This gives all integers from -100 to 200.</p><p><strong>Step 4:</strong> The overall range is from x^(-100+0) = x^(-100) to x^(101+200) = x^301. Since all integer powers in the range [-100, 301] appear at least once (by continuity of achievable powers through binomial expansions), the number of terms is 301-(-100)+1 = <strong>402</strong>.</p><p>∴ Answer: C</p>
Correct Answer: C