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7.Binomial Theorem
hard
यदि $\left(x^{\frac{1}{3}}+\frac{1}{2 x^{\frac{1}{3}}}\right)^{18},(x>0)$, के प्रसार में $x^{-2}$ तथा $x^{-4}$ के गुणांक क्रमशः $m$ तथा $n$ हैं, तो $\frac{m}{n}$ बराबर है
A
$27$
B
$182$
C
$\frac{5}{4}$
D
$\frac{4}{5}$
(JEE MAIN-2016)
Solution
$T_{r+1}=18 C_{r}\left(x^{\frac{1}{3}}\right)^{18-r}\left(\frac{1}{2 x^{\frac{1}{3}}}\right)^{r}$
$=^{18} C_{r} x^{6-\frac{2 r}{3}} \frac{1}{2^{r}}$
$\left\{ \begin{gathered}
6 – \frac{{2r}}{3} = – 2 \Rightarrow r = 12 \hfill \\
\& \,6 – \frac{{2r}}{3} = – 4 \Rightarrow r = 15 \hfill \\
\end{gathered} \right\}$
$\Rightarrow \quad \frac{\text { coefficient of } x^{-2}}{\text { coefficient of } x^{-4}}=\frac{^{18} C_{12} \frac{1}{2^{12}}}{^{18} C_{15} \frac{1}{2^{15}}}=182$
Standard 11
Mathematics