Find the middle terms in the expansion of:

(i) Here, n, i.e. 9, is an odd number.

Thus, the middle terms are $\left(\frac{n+1}{2}\right)$ th and $\left(\frac{n+1}{2}+1\right)$ th, i. e. 5 th and 6 th

Now,

$T_{5}=T_{4+1}={ }^{9} C_{4}(3 x)^{9-4}\left(\frac{-x^{3}}{6}\right)^{4}$

$=\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2} \times 27 \times 9 \times \frac{1}{36 \times 36} x^{17}$

$=\frac{189}{8} x^{17}$

And,

$T_{6}=T_{5+1}$

$={ }^{9} C_{5}(3 x)^{9-5}\left(\frac{-x^{3}}{6}\right)^{5}$

$=-\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2} \times 81 \times \frac{1}{216 \times 36} x^{19}$

$=-\frac{21}{16} x^{19}$

(ii) Here, n, i.e., 7, is an odd number.

Thus, the middle terms are $\left(\frac{7+1}{2}\right)$ th and $\left(\frac{7+1}{2}+1\right)$ th i. e. 4 th and 5 th

Now,

$T_{4}=T_{3+1}$

$={ }^{7} C_{3}\left(2 x^{2}\right)^{7-3}\left(\frac{-1}{x}\right)^{3}$

$=-\frac{7 \times 6 \times 5}{3 \times 2} \times 16 x^{8} \times \frac{1}{x^{3}}$

$=-560 x^{5}$

And,

$T_{5}=T_{4+1}$

$={ }^{7} C_{4}\left(2 x^{2}\right)^{7-4}\left(\frac{-1}{x}\right)^{4}$

$=35 \times 8 \times x^{6} \times \frac{1}{x^{4}}$

$=280 x^{2}$

(iii) Given :

$n$, i.e. 15 is an odd number.

Thus, the middle terms are $\left(\frac{15+1}{2}\right)$ th and $\left(\frac{15+1}{2}+1\right)$ th i.e. 8 th and 9 th.

Now,

$T_{8}=T_{7+1}$

$={ }^{15} C_{7}(3 x)^{15-7}\left(\frac{-2}{x^{2}}\right)^{7}$

$=-\frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9}{7 \times 6 \times 5 \times 4 \times 3 \times 2} \times 3^{8} \times 2^{7} x^{8-14}$

$=\frac{-6435 \times 3^{8} \times 2^{7}}{x^{6}}$

And,

$T_{9}=T_{8+1}$

$={ }^{15} C_{8}(3 x)^{15-8}\left(\frac{-2}{x^{2}}\right)^{8}$

$=\frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9}{7 \times 6 \times 5 \times 4 \times 3 \times 2} \times 3^{7} \times 2^{8} \times x^{7-16}$

$=\frac{6435 \times 3^{7} \times 2^{8}}{x^{9}}$

(iv) Here, n, i.e., 11, is an odd number.

Thus, the middle terms are $\left(\frac{11+1}{2}\right)$ th and $\left(\frac{11+1}{2}+1\right)$ th i. e. 6 th and 7 th.

Now,

$T_{6}=T_{5+1}$

$={ }^{11} C_{5}\left(x^{4}\right)^{11-5}\left(\frac{-1}{x^{3}}\right)^{5}$

$=-\frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2} \times(x)^{24-15}$

$=-462 x^{9}$

And,

$T_{7}=T_{6+1}$

$={ }^{11} C_{6}\left(x^{4}\right)^{11-6}\left(\frac{-1}{x^{3}}\right)^{6}$

$=\frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2}(x)^{20-18}$

$=462 x^{2}$