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(A) In the expansion of $(a+b)^{n}$,if $n$ is odd,there are two middle terms,namely the $\left(\frac{n+1}{2}\right)^{th}$ term and the $\left(\frac{n+

Vedclass · Accepted Answer

$-\frac{105}{8} x^{9}$ and $\frac{35}{48} x^{12}$

Vedclass · Answer

(A) In the expansion of $(a+b)^{n}$,if $n$ is odd,there are two middle terms,namely the $\left(\frac{n+1}{2}ight)^{th}$ term and the $\left(\frac{n+1}{2}+1ight)^{th}$ term.For the expansion $\left(3-\frac{x^{3}}{6}ight)^{7}$,$n=7$,so the middle terms are the $\left(\frac{7+1}{2}ight)^{th} = 4^{th}$ term and the $\left(\frac{7+1}{2}+1ight)^{th} = 5^{th}$ term.$T_{4} = T_{3+1} = {}^{7}C_{3} (3)^{7-3} \left(-\frac{x^{3}}{6}ight)^{3} = \frac{7 	imes 6 	imes 5}{3 	imes 2 	imes 1} 	imes 3^{4} 	imes \left(-\frac{x^{9}}{216}ight) = 35 	imes 81 	imes \left(-\frac{x^{9}}{216}ight) = -\frac{105}{8} x^{9}$.$T_{5} = T_{4+1} = {}^{7}C_{4} (3)^{7-4} \left(-\frac{x^{3}}{6}ight)^{4} = {}^{7}C_{3} 	imes 3^{3} 	imes \frac{x^{12}}{1296} = 35 	imes 27 	imes \frac{x^{12}}{1296} = \frac{35}{48} x^{12}$.Thus,the middle terms are $-\frac{105}{8} x^{9}$ and $\frac{35}{48} x^{12}$.

Find the middle terms in the expansions of $\left(3-\frac{x^{3}}{6}\right)^{7}$.

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