<p>The coefficient of \(x^{1012}\) in the expansion of \((1+x^n+x^{253})^{10}\), (where \(n\leq 22\) is any positive integer), is</p>
Step-by-Step Solution
Key Concept: Express (1+x^n+x^{253})^10 using multinomial theorem and find which combinations of powers sum to 1012. The constraint n≤22 is crucial—it determines which partitions of 1012 are achievable.
<p><strong>Step 1:</strong> Use multinomial theorem: (1+x^n+x^{253})^{10} = Σ [10!/(a!b!c!)] · 1^a · (x^n)^b · (x^{253})^c where a+b+c=10</p><p><strong>Step 2:</strong> For coefficient of x^{1012}, we need: nb + 253c = 1012 with a+b+c=10</p><p><strong>Step 3:</strong> Find valid combinations where n≤22:</p><ul><li>If c=4: 253(4)=1012, so nb=0 → b=0, a=6. Coefficient = 10!/(6!·0!·4!) = 210</li><li>If c=3: 253(3)=759, need nb=253. With n≤22, we'd need b≥12, but a+b+c=10 is impossible</li><li>If c≤2: 253(2)=506, need nb=506. Similar contradiction with a+b+c=10</li></ul><p><strong>Step 4:</strong> The only valid solution is a=6, b=0, c=4 (independent of n)</p><p>∴ Coefficient of x^{1012} = 210</p>
Correct Answer: B