The sum of the coefficients of x^{2/3} and x^ | vedclass

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Question

(A) The general term in the expansion of $(x^{2/3} + \frac{1}{2}x^{-2/5})^9$ is given by $T_{r+1} = {}^9C_r (x^{2/3})^{9-r} (\frac{1}{2} x^{-2/5})^r$.

Vedclass · Accepted Answer

$21/4$

Vedclass · Answer

(A) The general term in the expansion of $(x^{2/3} + \frac{1}{2}x^{-2/5})^9$ is given by $T_{r+1} = {}^9C_r (x^{2/3})^{9-r} (\frac{1}{2} x^{-2/5})^r$.$T_{r+1} = {}^9C_r (\frac{1}{2})^r x^{6 - \frac{2r}{3} - \frac{2r}{5}} = {}^9C_r (\frac{1}{2})^r x^{6 - \frac{16r}{15}}$.For the coefficient of $x^{2/3}$,set $6 - \frac{16r}{15} = \frac{2}{3}$.$6 - \frac{2}{3} = \frac{16r}{15}$ $\Rightarrow \frac{16}{3} = \frac{16r}{15}$ $\Rightarrow r = 5$.Coefficient of $x^{2/3} = {}^9C_5 (\frac{1}{2})^5 = 126 	imes \frac{1}{32} = \frac{63}{16}$.For the coefficient of $x^{-2/5}$,set $6 - \frac{16r}{15} = -\frac{2}{5}$.$6 + \frac{2}{5} = \frac{16r}{15}$ $\Rightarrow \frac{32}{5} = \frac{16r}{15}$ $\Rightarrow r = 6$.Coefficient of $x^{-2/5} = {}^9C_6 (\frac{1}{2})^6 = 84 	imes \frac{1}{64} = \frac{21}{16}$.Sum of coefficients $= \frac{63}{16} + \frac{21}{16} = \frac{84}{16} = \frac{21}{4}$.

The sum of the coefficients of $x^{2/3}$ and $x^{-2/5}$ in the binomial expansion of $(x^{2/3} + \frac{1}{2}x^{-2/5})^9$ is:

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