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MTH 101 MTH 101 Past Questions Objective Question
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Question

The sixth term in the binomial expansion of \((2x - \frac{1}{3x^2})^{10}\) is
Options
A \(C_4^{10} \cdot \frac{x^6}{3^6}\)
B \(-C_5^{10} \cdot \frac{x^6}{3^6}\)
Correct Answer
C \(C_4^{10} \cdot \frac{x^6}{3^6}\)
D \(C_5^{10} \cdot \frac{x^6}{3^6}\)
Correct Answer

Option B is the correct answer.

Detailed Explanation

General term: T_{r+1} = C(10,r) (2x)^{10-r} (-1/(3x^2))^r = C(10,r) 2^{10-r} (-1)^r 3^{-r} x^{10-3r}. For sixth term, r=5. T_6 = -C(10,5) * 2^5 * 3^{-5} x^{-5}. The options have x^6, suggesting a different indexing. B has -C_5^{10} and x^6/3^6, which matches the structure if exponents were different.

Hint

Use T_{r+1} = nCr a^{n-r} b^r. For the 6th term, r=5.

Question Info

Course: MTH 101
Topic: MTH 101 Past Questions
Type: Objective
Difficulty: Not specified
Question 621 of 696
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