Find the middle term in the expansionansion of `(3x-x^3/6)^9`.

Explanation

Here, `n`, i.e. 9 , is an odd number.

Thus, the middle terms are `(frac{n+1}{2})` th and `(frac{n+1}{2}+1) t h, i cdot e .5` th and 6 th

Now,

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

`=frac{9 times 8 times 7 times 6}{189 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} `

...