Question

Find the middle term in the expansion of ${(\frac{x}{y}-\frac{y}{x})}^7$.

Answer 1

${(\frac{x}{y}-\frac{y}{x})}^7$

$\because n=7\left(\text{odd}\right)$

There will be two middle terms, i.e.,

${\left(\frac{n+1}{2}\right)}^{th}$ term and ${\left(\frac{n+3}{2}\right)}^{th}$ term.

$\because n=7$

Middle term in the expansion ${(\frac{x}{y}-\frac{y}{x})}^7$ will be $4^{th}$ and $5^{th}$ terms.

As we know that general term of the expansion ${(a+b)}^n$ will be-

$T_{r+1}={}_n{C}_r{a}^{n-r}{b}^r$

Therefore,

$4^{th}$ term in the expansion ${(\frac{x}{y}-\frac{y}{x})}^7$-

$T_{3+1}={}_7{C}_3{\left(\frac{x}{y}\right)}^4{\left(\frac{y}{x}\right)}^3={}_7{C}_3\left(\frac{x^4}{y^4}\right)\left(\frac{y^3}{x^3}\right)={}_7{C}_3\frac{x}{y}$

$5^{th}$ term in the expansion ${(\frac{x}{y}-\frac{y}{x})}^7$-

$T_{4+1}={}_7{C}_4{\left(\frac{x}{y}\right)}^3{\left(\frac{y}{x}\right)}^4={}_7{C}_4\left(\frac{x^3}{y^3}\right)\left(\frac{y^4}{x^4}\right)={}_7{C}_4\frac{y}{x}$

Hence the middle term of the expansion ${(\frac{x}{y}-\frac{y}{x})}^7$ are-

$T_4={}_7{C}_3\frac{x}{y}$ and $T_5={}_7{C}_4\frac{y}{x}$