For a general $n$, the constant term in $$f(x) = (1 + x/5)^n (2-3/x)^2$$ can be found by observing first that $$(2-3/x)^2 = 2 - 12x^{-1} + 9x^{-2}.$$ Then we see that the constant term of $f$ is given by $$1 \cdot 2 + \binom{n}{1} (x/5)(-12x^{-1}) + \binom{n}{2}(x/5)^2(9x^{-2}),$$ since these are the only terms for which the power of $x$ will be zero.
This is probably the wrong proof for you, but I will post it anyways. (requires calculus)
Note that $f(x)=(a+x)^n$ is an analytic function in $x$ for arbitrary $a,n$ since on its own, it is a power series with one term.
If it is an analytic function, then it should follow Taylor's theorem.
Now, if we take the expansion around $x=0$, we get
$$(a+x)^n=a^n+na^{n-1}x+\frac{n(n+1)}2a^{n-2}x^2+\dots$$
Since $f(0)=a^n$, $f'(0)=na^{n-1}$, $\dots f^{(k)}(0)=n(n+1)(n+2)\dots(n+k-1)a^{n-k}$
or
$$(a+x)^n=\sum_{k=0}^\infty\frac{n(n+1)(n+2)\dots(n+k-1)}{k!}a^{n-k}x^n$$
$$(a+x)^n=\sum_{k=0}^\infty\binom nka^{n-k}x^n$$
where $f'(x)$ is the first derivative of $f(x)$, $f''(x)$ the second derivative, etc. $f^{(k)}(x)$ is the $k$th derivative of $f(x)$.
Best Answer
There is not an expansion for this by the usual binomial expansion. However, there is one by the generalized binomial theorem, which gives, for $n \in \Bbb Z^+$,
$$(1-x)^{-n} = \sum_{k=0}^\infty \binom{n+k-1}{n-1} x^k$$
Factor out $a$ from your expression and replace adding $b$ with subtracting $(-b)$:
$$(a+b)^{-1} = a^{-1} \left(1- \left(-\frac b a \right) \right)^{-1}$$
Now apply the theorem for $x=-b/a$ and $n=1$.
Footnote:
You could approach this by the geometric series instead, but this generalizes it somewhat. I used the above solution because "binomial" was specified, but geometric is easier. In the geometric series, we have
$$(1-x)^{-1} = \frac{1}{1-x} = \sum_{k=0}^\infty x^k$$
Do the factoring as before and take $x=-b/a$ for the same result.