Question

Find the middle terms in the expansion of ${(\frac{x}{3}+9y)}^{10}$.

Answer 1

Here n=10, which is even.

So the middle term is $(\frac{10}{2}+1)$th i.e., 6th term.

The general term in the expansion of ${(\frac{x}{3}+9y)}^{10}$ is

$T_{r+1}={10}_{C_r}{\left(\frac{x}{3}\right)}^{10-r}.(9y{)}^r$ ...(i)

Putting r=5 in (i)

$T_6={10}_{C_5}{\left(\frac{x}{3}\right)}^{10-5}.(9y{)}^5$

$={10}_{C_5}\frac{x^5}{3^5}{.9}^5.y^5$

$=252\times \frac{x^5}{243}\times 59049y^5$

$=61236x^5{y}^5$