Question

If the constant term of the binomial expansion ${(2x-\frac{1}{x})}^n$ is $-160$, then $n$ is equal to -

• A. $4$
• B. $6$
• C. $8$
• D. $10$

Answer 1

Answer: (B)

$6$

General term of $(2x-\frac{1}{x}{)}^n$ is ${}^n{C}_r\left(2x{)}^{n-r}\right(\frac{-1}{x}{)}^r$

for constant term $n-r=r⟹r=\frac{n}{2}$

the constant term is $^{n}C_{\dfrac{n}{2}}(2)^{\dfrac{n}{2}}(-1)^{\dfrac{n}

{2}}=-160$

$⟹n=6$