AP EAMCET · Maths · Binomial Theorem
The coefficient of \(x^f\) in the expansion of \(\frac{1}{\sqrt[3]{(1-2 x)^2}}\) is
- A \(\frac{2.5 .8 \ldots(3 r-1)}{r !}(-1)^r\left(\frac{2}{3}\right)^r\)
- B \(\frac{2.5 .8 \ldots(3 r-1)}{r !}(-1)^r\left(\frac{3}{2}\right)^r\)
- C \(\frac{2.5 \cdot 8 \ldots(3 r-1)}{r !}\left(\frac{2}{3}\right)^r\)
- D \(\frac{2.5 .8 \ldots(3 \mathrm{r}-1)}{\mathrm{r} !}\left(\frac{3}{2}\right)^{\mathrm{r}}\)
Answer & Solution
Correct Answer
(C) \(\frac{2.5 \cdot 8 \ldots(3 r-1)}{r !}\left(\frac{2}{3}\right)^r\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{\sqrt[3]{(1-2 x)^2}}=\frac{1}{(1-2 x)^{2 / 3}}=(1-2 x)^{-2 / 3}\) \(=1+\left(\frac{2}{3}\right)(2 x)+\frac{\frac{2}{3}\left(\frac{2}{3}+1\right)}{2 !}(2 x)^2\) \(+\frac{\frac{2}{3}\left(\frac{2}{3}+1\right)\left(\frac{2}{3}+2\right)}{3 !}(2 x)^3+\ldots \ldots\)…
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