Question

The middle term in the expansion of $(1-\frac{1}{x}{)}^{10}$ is :

Answer 1

Total number of terms$=10+1=11$

To find :$r+1=6$

$r=5$

Given:$n=10,a=1,b=\frac{-1}{x}$

We have Middle term${=}^n{C}_r{a}^{n-r}{b}^r$

${=}^{10}{C}_5{1}^{10-5}{\left(\frac{-1}{x}\right)}^5$

$=\frac{-252}{x^5}$

$\therefore$ the middle term$=\frac{-252}{x^5}$