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. The sum of coefficients is found by setting x=1.
<p><strong>Step 1:</strong> Simplify the given expression. Notice that 1+2x+x² = (1+x)², so:</p><p>(1+2x+x²)ⁿ = [(1+x)²]ⁿ = (1+x)²ⁿ</p><p><strong>Step 2:</strong> Therefore:</p><p>(1+x)²ⁿ + (1+x)²ⁿ = 2(1+x)²ⁿ = Σ aᵣxʳ</p><p><strong>Step 3:</strong> To find Σ aᵣ (sum of all coefficients), substitute x=1:</p><p>Σ aᵣ = 2(1+1)²ⁿ = 2(2)²ⁿ = 2·2²ⁿ = 2²ⁿ⁺¹</p><p><strong>Step 4:</strong> Therefore f(n) = 2²ⁿ⁺¹ = 4ⁿ·2</p><p>∴ Answer: BC (matches options containing f(n) = 2²ⁿ⁺¹ or equivalent form)</p>
Correct Answer: BC