Question

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

Answer 1

Compute the middle term.

We know that the general form of ${\left(a+b\right)}^n$ is $T_{r+1}={}^{n}C_r{a}^{n-r}{b}^r$.

Here $n=10,a=\frac{x}{3}.b=9y$.

Since $n$ is even,

The middle term $=\left(\frac{n}{2}+1\right)$th term.

$=\frac{10}{2}+1$th term

$=6$th term

Here $T_6=T_{5+1}$. That is,

$r=5$

So we can write that

$T_{5+1}={}^{10}C_5{\left(\frac{x}{3}\right)}^{10-5}{\left(9y\right)}^5$

$$\begin{gathered} ={}^{10}C_5{\left(\frac{x}{3}\right)}^5{\left(9\right)}^5\left(y^5\right) \\ =\frac{10!}{5!\left(10-5\right)!}{\left(\frac{x}{3}\right)}^5{\left(3^2\right)}^5\left(y^5\right) \\ =\frac{10!}{5!5!}\times \frac{x^5}{3^5}\times 3^{10}\times y^5 \end{gathered}$$ $\left[\therefore {}^{n}C_r=\frac{n!}{(n–r)!r!}\right]$

$$\begin{gathered} =\frac{10\times 9\times 8\times 7\times 6\times 5!}{5\times 4\times 3\times 2\times 1\times 5!}\times x^5\times 3^5\times y^5 \\ =\frac{10\times 9\times 8\times 7\times 6\times 3^5}{5\times 4\times 3\times 2\times 1}\times x^5\times y^5 \\ =61236x^5{y}^5 \end{gathered}$$

Hence, the middle term of the expansion is $\mathbf{61236}\mathbf{}{\mathit{x}}^{\mathbf{5}}{\mathit{y}}^{\mathbf{5}}$.