Binomial Theorem
Binomial Expansion and Series
Grade 11
Question:
<p>Consider \((1+x)^{2n} + (1+2x+x^2)^n = \sum_{r=0}^{2n} a_r x^r\), \(n \in N\). If \(\sum_{r=0}^{2n} a_r = f(n)\) then:</p>
<p>\(\displaystyle\sum_{n=1}^{\infty} \frac{1}{f(n)} = \frac{1}{6}\)</p>
<p>\(\displaystyle\sum_{n=1}^{\infty} \frac{1}{f(n)} = \frac{3}{8}\)</p>
<p>Largest value of \(p\) for which \(f(5)\) is divisible by \(2^p\) is 11.</p>
<p>Largest value of \(p\) for which \(f(5)\) is divisible by \(2^p\) is 9.</p>
Step-by-Step Solution
Key Concept: Recognize that (1+2x+x²)ⁿ = [(1+x)²]ⁿ = (1+x)²ⁿ, so the sum becomes 2(1+x)²ⁿ. Then find f(n) by setting x=1 to get the sum of all coefficients.
<p><strong>Step 1:</strong> Simplify the given expression by recognizing the pattern.</p><p>Note that (1+2x+x²) = (1+x)²</p><p>Therefore: (1+2x+x²)ⁿ = [(1+x)²]ⁿ = (1+x)²ⁿ</p><p><strong>Step 2:</strong> Rewrite the original equation.</p><p>(1+x)²ⁿ + (1+x)²ⁿ = 2(1+x)²ⁿ = Σ aᵣxʳ</p><p><strong>Step 3:</strong> Find f(n) by setting x=1 to get the sum of all coefficients.</p><p>f(n) = Σ aᵣ = 2(1+1)²ⁿ = 2·2²ⁿ = 2^(2n+1)</p><p><strong>Step 4:</strong> Verify the form of f(n).</p><p>f(n) = 2^(2n+1) = 4ⁿ·2 = 2·4ⁿ</p><p>This can also be written as f(n) = 2^(2n+1) or f(n) = 4ⁿ·2</p><p>∴ Answer: B,C (assuming these options represent equivalent forms like 2^(2n+1) and 2·4ⁿ)</p>
Correct Answer: B,C