# Find the middle terms in the expansion of    Q7.    $\left(3 - \frac{x^3}{6} \right )^7$

As we know that the middle  terms in the expansion of  $(a+b)^n$ when n is odd are,

$\left ( \frac{n+1}{2} \right )^{th}\:term\: \:and\:\:\left ( \frac{n+1}{2}+1 \right )^{th}\:term$

Hence the middle term of the expansion    $\left(3 - \frac{x^3}{6} \right )^7$   are

$\left ( \frac{7+1}{2} \right )^{th}\:term\: \:and\:\:\left ( \frac{7+1}{2}+1 \right )^{th}\:term$

Which are $4^{th}\:term\:and\:\:5^{th} \:term$

Now,

As we know that the general  $(r+1)^{th}$ term  $T_{r+1}$ in the binomial expansion of  $(a+b)^n$  is given by

$T_{r+1}=^nC_ra^{n-r}b^r$

So the $4^{th}$ term of the expansion of   $\left(3 - \frac{x^3}{6} \right )^7$ is

$\\\Rightarrow T_4= T_{3+1}\\=^{7}C_3(3)^{7-3}\left (- \frac{x^3}{6} \right )^3\\=\left ( -\frac{1}{6} \right )^3\times 3^4\times^{7}C_3\times x^{9}\\=\left ( -\frac{1}{6} \right )^3\times 3^4\times\frac{7!}{3!4!}\times x^{9} \\=-\frac{3\times3\times3\times3}{6\times6\times6}\times\frac{7\times 6\times5}{3\times2}\times x^9$

$=-\frac{105}{8}x^9$

And the $5^{th}$ Term of the expansion of   $\left(3 - \frac{x^3}{6} \right )^7$  is,

$\\\Rightarrow T_5= T_{4+1}\\=^{7}C_4(3)^{7-4}\left (- \frac{x^3}{6} \right )^4\\=\left ( -\frac{1}{6} \right )^4\times 3^3\times^{7}C_4\times x^{12}\\=\left ( -\frac{1}{6} \right )^4\times 3^3\times\frac{7!}{3!4!}\times x^{12} \\=\frac{3\times3\times3}{6\times6\times6\times6}\times\frac{7\times 6\times5}{3\times2}\times x^{12}$

$=\frac{35}{48}x^{12}$

Hence the middle terms of the expansion of given expression are

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

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