AP EAMCET · Maths · Binomial Theorem
If \((2-5 x)^{\frac{-1}{5}}=a_0+a_1 x+a_2 x^2+\ldots\), then \(\frac{a_1}{a_2}=\)
- A \(\frac{1}{3}\)
- B \(-\frac{2}{3}\)
- C \(-\frac{1}{3}\)
- D \(\frac{2}{3}\)
Answer & Solution
Correct Answer
(D) \(\frac{2}{3}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { }(2-5 x)^{-1 / 5}=(2)^{-1 / 5}\left(1-\frac{5}{2} x\right)^{-1 / 5} \\ & =(2)^{-1 / 5}\left[1+\frac{1}{2} x+\frac{3}{4} x^2+\ldots\right] \\ & =(2)^{-1 / 5}+\frac{1}{2} \cdot(2)^{-1 / 5} x+\frac{3}{4}(2)^{-1 / 5} x^2+\ldots \\ & \text { Hence }…
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