Question

If the middle term in the expansion of ${(x+\frac{b}{x})}^6$ is $160$, find $b$.

Answer 1

${(x+\frac{b}{x})}^6$ Total term$=7$

middle term=$\frac{7+1}{2}=\frac{8}{2}=4^{th}$

$T_4$ ${=}^6{C}_3$ $x^3$ ${\left(\frac{b}{x}\right)}^3$

$160=\frac{6!}{3!3!}.x^3\frac{b^3}{x^3}$

$b^3=8$

$\Rightarrow b=2$.