AP EAMCET · Maths · Binomial Theorem
The coefficient of \(x^5\) in the expansion of \(\left(2 x^3-\frac{1}{3 x^2}\right)^5\) is
- A \(8\)
- B \(9\)
- C \(\frac{80}{9}\)
- D \(\frac{29}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{80}{9}\)
Step-by-step Solution
Detailed explanation
General term of \(\left(2 x^3-\frac{1}{3 x^2}\right)^5\) is \(T_{r+1}={ }^5 C_r\left(2 x^3\right)^{5-r}\left(\frac{-1}{3 x^2}\right)^r\) \(=(2)^{5-r}\left(\frac{-1}{3}\right)^r{ }^5 C_r x^{15-3 r-2 r}\) \(=(2)^{5-r}\left(\frac{-1}{3}\right)^r{ }^5 C_r x^{15-5 r}\) For…
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