Binomial Theorem
Terms in Binomial Expansion
Grade 11
Question:
<p>A ratio of the 5<sup>th</sup> term from the beginning to the 5<sup>th</sup> term from the end in the binomial expansion of \(\left(2^{1/3} + \dfrac{1}{2(3)^{1/3}}\right)^{10}\) is:</p>
<p>\(1 : 2(6)^{\frac{1}{3}}\)</p>
<p>\(1 : 4(16)^{\frac{1}{3}}\)</p>
<p>\(4(36)^{\frac{1}{3}} : 1\)</p>
<p>\(2(36)^{\frac{1}{3}} : 1\)</p>
Step-by-Step Solution
Key Concept: In a binomial expansion of (a+b)^n, the 5th term from the end equals the (n-4)th term from the beginning. Use the symmetry property: T_(r) from beginning = T_(n-r+2) from end. Here n=10, so 5th from end = 6th from beginning.
<p><strong>Step 1:</strong> Identify positions. In (a+b)^10, the 5th term from beginning is T₅ and 5th term from end is T₆ (since total terms = 11).</p><p><strong>Step 2:</strong> Find T₅: T₅ = C(10,4)·(2^(1/3))^6·(1/(2·3^(1/3)))^4</p><p>= C(10,4)·2²·1/(2^4·3^(4/3)) = 210·4·1/(16·3^(4/3)) = 210/(4·3^(4/3))</p><p><strong>Step 3:</strong> Find T₆: T₆ = C(10,5)·(2^(1/3))^5·(1/(2·3^(1/3)))^5</p><p>= C(10,5)·2^(5/3)·1/(2^5·3^(5/3)) = 252·2^(5/3)/(32·3^(5/3)) = 252/(2^(10/3)·3^(5/3))</p><p><strong>Step 4:</strong> Calculate ratio T₅/T₆:</p><p>= [210/(4·3^(4/3))] / [252/(2^(10/3)·3^(5/3))]</p><p>= [210·2^(10/3)·3^(5/3)] / [4·3^(4/3)·252]</p><p>= [210·2^(10/3)·3^(1/3)] / [4·252] = [210·2^(10/3)·3^(1/3)] / 1008</p><p>= [2^(10/3)·3^(1/3)] / 4.8 = 2^(4/3)·3^(1/3) / 2 = 2^(1/3)·3^(1/3) = (6)^(1/3) or <strong>∛6</strong></p><p>∴ Answer: D</p>
Correct Answer: D