The middle terms in the expansion of {left( { | vedclass

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Question

(A) The binomial expression is ${\left( {3x - \frac{{{x^3}}}{6}} \right)^9}$.Here,$n = 9$,which is odd,so there are two middle terms: the $\left( \fra

Vedclass · Accepted Answer

$-\frac{21}{16}x^{19}, \frac{189}{8}x^{17}$

Vedclass · Answer

(A) The binomial expression is ${\left( {3x - \frac{{{x^3}}}{6}} ight)^9}$.Here,$n = 9$,which is odd,so there are two middle terms: the $\left( \frac{n+1}{2} ight)^{th}$ and $\left( \frac{n+3}{2} ight)^{th}$ terms.These are the $5^{th}$ and $6^{th}$ terms.The general term is given by $T_{r+1} = {}^nC_r (a)^{n-r} (b)^r$.For $r=4$ ($5^{th}$ term): $T_5 = {}^9C_4 (3x)^{9-4} (-\frac{x^3}{6})^4 = 126 	imes (3x)^5 	imes \frac{x^{12}}{6^4} = 126 	imes 243x^5 	imes \frac{x^{12}}{1296} = \frac{126 	imes 243}{1296} x^{17} = \frac{189}{8} x^{17}$.For $r=5$ ($6^{th}$ term): $T_6 = {}^9C_5 (3x)^{9-5} (-\frac{x^3}{6})^5 = 126 	imes (3x)^4 	imes (-\frac{x^{15}}{6^5}) = 126 	imes 81x^4 	imes (-\frac{x^{15}}{7776}) = -\frac{126 	imes 81}{7776} x^{19} = -\frac{21}{16} x^{19}$.Thus,the middle terms are $-\frac{21}{16}x^{19}$ and $\frac{189}{8}x^{17}$.

The middle terms in the expansion of ${\left( {3x - \frac{{{x^3}}}{6}} \right)^9}$ are:

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