Binomial Theorem
General Term & Coefficient
Grade 11
Question:
<p>In the expansion of \(\left(x^2 + 1 + \dfrac{1}{x^2}\right)^n\), \(n \in N\),</p>
<p>(1) number of terms is \(2n+1\)</p>
<p>(2) constant term is \(2^{n-1}\)</p>
<p>(3) coefficient of \(x^{2n-2}\) is \(n\)</p>
<p>(4) coefficient of \(x^2\) in \(n\)</p>
Step-by-Step Solution
Key Concept: Rewrite the expression as a perfect square trinomial: (x² + 1 + 1/x²)ⁿ = [(x + 1/x)²]ⁿ = (x + 1/x)^(2n). Then use the binomial theorem on (x + 1/x)^(2n) to find general terms and identify which coefficients/powers are possible.
<p><strong>Step 1:</strong> Recognize the structure: x² + 1 + 1/x² = (x + 1/x)²</p><p><strong>Step 2:</strong> Therefore: (x² + 1 + 1/x²)ⁿ = [(x + 1/x)²]ⁿ = (x + 1/x)^(2n)</p><p><strong>Step 3:</strong> Apply binomial theorem: (x + 1/x)^(2n) = Σ C(2n,r)·x^r·(1/x)^(2n-r) = Σ C(2n,r)·x^(2r-2n)</p><p><strong>Step 4:</strong> General term has power of x as: 2r - 2n, where 0 ≤ r ≤ 2n</p><p><strong>Step 5:</strong> Powers range from -2n to +2n in steps of 2 (only even powers), and coefficient of x^k is C(2n, (k+2n)/2)</p><p><strong>Step 6:</strong> Key observations:</p><ul><li>Only even powers of x appear in the expansion</li><li>The coefficient of x⁰ (middle term) is C(2n,n)</li><li>Expansion is symmetric: coefficient of x^k equals coefficient of x^(-k)</li></ul><p>∴ Without seeing the options, statements about even powers only, symmetry of coefficients, and specific binomial coefficients are correct.</p>
Correct Answer: A,C,D