Binomial Theorem
Sum of binomial coefficients with integral
Grade 11
Question:
<p>The value of \(\dfrac{{}^nC_0}{n} + \dfrac{{}^nC_1}{n+1} + \dfrac{{}^nC_2}{n+2} + \cdots + \dfrac{{}^nC_n}{2n}\) is equal to</p>
<p>(1) \(\displaystyle\int_0^1 x^{n-1}(1-x)^n\,dx\)</p>
<p>(2) \(\displaystyle\int_1^2 x^n(x-1)^{n-1}\,dx\)</p>
<p>(3) \(\displaystyle\int_1^2 x^{n-1}(1+x)^n\,dx\)</p>
<p>(4) \(\displaystyle\int_0^1 (1-x)^n x^{n-1}\,dx\)</p>
Step-by-Step Solution
Key Concept: Recognize that ∫₀¹ xʳ dx = 1/(r+1), allowing you to express each term ⁿCᵣ/(n+r) as the integral ⁿCᵣ∫₀¹ xⁿ⁺ʳ dx. Then sum using the binomial theorem: (1+x)ⁿ = Σ ⁿCᵣxʳ.
<p><strong>Step 1: Recognize the integral form</strong></p><p>Note that 1/(n+r) = ∫₀¹ xⁿ⁺ʳ⁻¹ dx, so:</p><p>ⁿCᵣ/(n+r) = ⁿCᵣ ∫₀¹ xⁿ⁺ʳ⁻¹ dx</p><p><strong>Step 2: Sum all terms</strong></p><p>∑ᵣ₌₀ⁿ [ⁿCᵣ/(n+r)] = ∑ᵣ₌₀ⁿ ⁿCᵣ ∫₀¹ xⁿ⁺ʳ⁻¹ dx</p><p>= ∫₀¹ xⁿ⁻¹ ∑ᵣ₌₀ⁿ ⁿCᵣ xʳ dx</p><p><strong>Step 3: Apply binomial theorem</strong></p><p>= ∫₀¹ xⁿ⁻¹(1+x)ⁿ dx</p><p><strong>Step 4: Evaluate the integral</strong></p><p>Let u = 1+x, then du = dx. When x=0, u=1; when x=1, u=2:</p><p>= ∫₁² (u-1)ⁿ⁻¹ uⁿ du</p><p>Using the beta function or integration by parts repeatedly:</p><p>= ∫₀¹ xⁿ⁻¹(1+x)ⁿ dx = (2ⁿ⁺¹ - 1)/(n+1)</p><p><strong>∴ Answer: (2ⁿ⁺¹ - 1)/(n+1)</strong></p>
Correct Answer: A