Question Detail

MTH 101 MTH 101 Past Questions Objective Question

Question

The sixth term in the binomial expansion of \( (2x - \frac{1}{3x^2})^{10} \) is not equal to 0 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

The general term in the expansion of (a + b)^n is T_{r+1} = C(n, r) a^{n-r} b^r. Here, a = 2x, b = -1/(3x^2), n=10. The sixth term corresponds to r=5 (since r+1=6). T_6 = C(10,5) (2x)^{10-5} * (-1/(3x^2))^5 = C(10,5) (2^5 x^5) * (-1)^5 / (3^5 x^{10}) = - C(10,5) * 32 / 243 * x^{5-10} = - C(10,5) * 32 / 243 * x^{-5}. The options have x^6, which suggests a different indexing or a typo. Option B has -C_5^{10} * x^6/3^6. If we ignore the coefficient 2^5 and the x^{-5} vs x^6 discrepancy, the structure with the negative sign and C_5^{10} matches. Option B is selected as the most likely intended answer.

Hint

Remember the general term formula: T_{r+1} = nCr * a^{n-r} * b^r. For the 6th term, r=5. Pay attention to signs and exponents.

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