Question

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

Answer 1

Here n= 7, which is odd.

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

The general term in the expansion of ${(3-\frac{x^3}{6})}^7$ is

$T_{r+1}=7_{C_r}(3{)}^{7-r}{\left(\frac{-x^3}{6}\right)}^r$ ...(i)

Putting r=3 and 4 in (i)

$\therefore T_4=7_{C_3}(3{)}^{7-3}{(-\frac{x^3}{6})}^3$

$=7_{C_3}(3{)}^4.(-1{)}^3.\frac{x^9}{(6{)}^3}$

$=35\times 81\times -\frac{x^9}{216}=\frac{-105}{8}{x}^9$

Now, $T_5=7_{C_4}(3{)}^{7-4}{\left(\frac{-x^3}{6}\right)}^4$

$=7_{C_4}(3{)}^3.(-1{)}^4\frac{x^{12}}{(6{)}^4}$

$=35\times 27\times \frac{x^{12}}{1296}=\frac{35}{48}{x}^{12}$