Maclaurin series and general term for $e^{(x-2)^2}$

Question

I need to write the first four nonzero terms and the general term for the Taylor series for $g(x)=e^{(x-2)^2}$ about $x=2$. I tried foiling out $(x-2)^2$ and using that whole term to replace $x$ in the general term and in the series, but it gets very messy and I do not believe I am correct. Thanks in advance!

Answer 1

It does get ugly, true...

$\begin{align*} \exp((x - 2)^2) &= \sum_{n \ge 0} \frac{(x - 2)^{2 n}}{n!} \\ &= \sum_{n \ge 0} \frac{1}{n!} \sum_{0 \le k \le 2 n} \binom{2 n}{k} \cdot (-2)^{2 n - k} x^k \\ &= \sum_{n \ge 0} \sum_k \frac{1}{n!} \binom{2 n}{k} \cdot (-2)^{2 n - k} x^k \\ &= \sum_k x^k \sum_{n \ge 0} \frac{1}{n!} \binom{2 n}{k} \cdot (-2)^{2 n - k} \end{align*}$

We can drop the range of the inner sum as outside the range the binomial coefficient is zero, and switching summation orders around is permissible as long as the mess converges absolutely, which it does for the exponential sum and the internal sum is finite.

A "nice" form for the coefficient (inner sum) is rather unlikely.