If the constant term of binomial expansion $ (2x-\frac{1}{x})^{n}$ is$-160$ ,then $n$ equal to? As I know the expansion binomial expression of $ (2x-\frac{1}{x})^{n}$ .But I don't know that for which $-160$ in $n$th term of given expression.
Please help me to solve this.
 A: As usual, the binomial expansion helps:
$$
\left(2x - \frac 1x\right)^n = \sum_{k=0}^n (-1)^k\binom nk\frac{1}{x^k}(2x)^{n-k} = \sum_{k=0}^n \binom nk (-1)^k2^{n-k}x^{n-2k}
$$
Now, when does the constant term appear? Precisely when $n=2k$, as one sees above. Also, if $n$ is odd, then you see that the constant term is zero, so $n$ must be even.
Then, the constant itself is $(-1)^k\binom nk 2^{n -k } = \binom n{\frac n2} 2^{\frac n2}$.
Now, this is equal to $-160$, as you say. We note that $160$ is a multiple of $5$, hence $n \geq 5$ must happen, otherwise $\binom nk$ cannot be a multiple of $5$. Also, $k$ is odd, as the coefficient is negative, hence $n = 6,10,...$ As we notice, $n=6$ gives $\binom 63 \times 8 = 20 \times 8 = 160$. Hence, the answer is $n = 6$.
Almost as if to prove a point:

$ \left(2x - \frac 1x\right)^6 = 64 x^6 + \frac 1{x^6} - 192 x^4 - \frac{12}{x^4} + 240 x^2 + \frac{60}{x^2} - 160$.

A: As remarked, $n$ is even, i.e., $n=2k$. Hence the "constant" term of $(2x-1/x)^{2k}$ is 
$$
-160=\binom{2k}{k}(-2)^k.
$$
(In particular, if such $k$ exists, then it is odd.) Note the function on the right hand side is strictly increasing in $k$ hence $k=3$ is the unique solution, i.e., $n=6$.
