If the coefficients of x-2 and x-4 in the expansion of x1-3-1-2 x1-318-x-0- are m and n respectively- then m-n is equal to -

The correct option is B 182T_r+1=^18C_r (x^1/3)^18 r(1/2x^1/3)^r =^18C_r (1/2)^r x^(18 2r/3) For coefficient of x^ 2, nbsp;18 2r/3= 2 ⇒ r=12 For coeffici ...

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Byju's Answer

Standard XII

Mathematics

General Term of Binomial Expansion

If the coeffi...

Question

If the coefficients of $x^{- 2}$ and $x^{- 4}$ in the expansion of ${(x^{\frac{1}{3}} + \frac{1}{2 x^{\frac{1}{3}}})}^{18}, (x > 0),$ are $m$ and $n$ respectively, then $\frac{m}{n}$ is equal to :

\frac{4}{5}

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182

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27

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\frac{5}{4}

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Solution

The correct option is B $182$
$T_{r + 1} =^{18} C_{r} (x^{\frac{1}{3}})^{18 - r} {(\frac{1}{2 x^{\frac{1}{3}}})}^{r} =^{18} C_{r} {(\frac{1}{2})}^{r} x^{(\frac{18 - 2 r}{3})} For coefficient of x^{- 2}, \frac{18 - 2 r}{3} = - 2 \Rightarrow r = 12 For coefficient of x^{- 4}, \frac{18 - 2 r}{3} = - 4 \Rightarrow r = 15 \frac{m}{n} = \frac{^{18} C_{12} {(\frac{1}{2})}^{12}}{^{18} C_{15} {(\frac{1}{2})}^{15}} = 182$

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If the coefficients of x–2 and x–4 in the expansion of (x13+12x13)18,(x>0), are m and n respectively, then mn is equal to :

The correct option is B 182Tr+1=18Cr(x13)18−r(12x13)r=18Cr(12)rx(18−2r3)For coefficient of x−2,18−2r3=−2⇒r=12For coefficient of x−4,18−2r3=−4⇒r=15mn=18C12(12)1218C15(12)15=182

If the coefficients of $x^{- 2}$ and $x^{- 4}$ in the expansion of ${(x^{\frac{1}{3}} + \frac{1}{2 x^{\frac{1}{3}}})}^{18}, (x > 0),$ are $m$ and $n$ respectively, then $\frac{m}{n}$ is equal to :