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
You could calculate, for example, $(1+x)^{1/2}=a_0+a_1x+a_2x^2+\cdots$ by squaring both sides and comparing coefficients. For example we can get the first three coefficients by ignoring all degree $3$ terms and higher: $$1+x=a_0^2+2a_0a_1x+2a_0a_2x^2+a_1^2x^2+\cdots$$
From here we can conclude that $a_0=\pm1$ (we'll take $+1$ to match what happens when $x=0$). Then comparing coefficients of $x$ we have $2a_1=1$, so $a_1=1/2$. Finally, comparing coefficients of $x^2$, we have $2a_0a_2+a_1^2=0$, so $2a_2+1/4=0$ and $a_2=-1/8$.
You can definitely get as many coefficients as you want this way, and I trust that you can even derive the binomial coefficient formula. However, this is not any easier than the Taylor series, where you take $(1+x)^{1/2}=a_0+a_1x+a_2x^{2}+\cdots$ and find the coefficients by saying the $n$th derivatives on both sides have to be equal at $0$.
For example, plugging in $0$ on both sides we conclude $a_0=1$. Calculating the first derivative of both sides, we have $$\frac{1}{2}(x+1)^{-1/2}=a_1+2a_2x+\cdots$$ Plugging in $0$, we get $a_1=1/2$. Taking the derivative one more time, we see $$(-1/2)(1/2)(1+x)^{-3/2}=2a_2+\cdots$$ Plugging in $x=0$, we have $(-1/2)(1/2)=2a_2$, or $a_2=-1/8$. The advantage to this way, is that it is much easier to see the pattern of coefficients!
Unfortunately, there is a big hole in both arguments. They will give you what the coefficients have to be, but they won't prove that the series expansion converges in the first place. We started off by assuming you could write $1+x$ as an infinite power series, but there is no guarentee that this exists, and actually it doesn't converge unless $|x|<1$, which we never used. So you need to estimate the error in Taylor's formula to complete the proof rigorously.