Binomial Expansion Question

Question

I have $(2x-\frac{1}{2x})^6$. To calculate the constant term I used the term formula and found the value of $r$ which would give me the constant term, which I found to be 3. According to the Maclaurin series, isn't the coefficient of $x^0$, the constant term, given by $f(0)$? I used that here and I'm getting different answers. Why?

Answer 1

The function is not defined at $x=0$, indeed blows up there. So the Maclaurin series does not exist.

The coefficient of $x^0$ is obtained by expanding $(a+b)^6$ using the Binomial Theorem, where $a=2x$ and $b=-\frac{1}{2x}$. We get the $x^0$ term from $\binom{6}{3}a^3b^3$.

Calculate. We get $\binom{6}{3}(2x)^3(-1/(2x))^3=-20x^0$.