The sum $$\sum_{n=1}^{\infty} \frac{2n^2+3n+4}{(2n)!}$$ is equal to:

We need to find $$\sum_{n=1}^{\infty} \frac{2n^2+3n+4}{(2n)!}$$.

First, we write the series as $$S = 2\sum_{n=1}^{\infty}\frac{n^2}{(2n)!} + 3\sum_{n=1}^{\infty}\frac{n}{(2n)!} + 4\sum_{n=1}^{\infty}\frac{1}{(2n)!}$$.

Recall that $$\cosh(1) = \frac{e + e^{-1}}{2} = \sum_{n=0}^{\infty} \frac{1}{(2n)!}$$ and $$\sinh(1) = \frac{e - e^{-1}}{2} = \sum_{n=0}^{\infty} \frac{1}{(2n+1)!}$$.

Since $$\sum_{n=1}^{\infty}\frac{1}{(2n)!} = \cosh(1) - 1 = \frac{e + e^{-1}}{2} - 1$$, we have this third component.

Using $$\frac{n}{(2n)!} = \frac{1}{2} \cdot \frac{1}{(2n-1)!}$$, it follows that $$\sum_{n=1}^{\infty}\frac{n}{(2n)!} = \frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{(2n-1)!} = \frac{1}{2}\sinh(1) = \frac{e - e^{-1}}{4}$$.

To evaluate the term with $$n^2$$ we write $$n^2 = \frac{n(2n-1)}{2} + \frac{n}{2}$$ so that $$\frac{n^2}{(2n)!} = \frac{1}{4} \cdot \frac{1}{(2n-2)!} + \frac{1}{4} \cdot \frac{1}{(2n-1)!}$$.

Hence $$\sum_{n=1}^{\infty}\frac{n^2}{(2n)!} = \frac{1}{4}\sum_{k=0}^{\infty}\frac{1}{(2k)!} + \frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{(2n-1)!} = \frac{1}{4}\cosh(1) + \frac{1}{4}\sinh(1) = \frac{e}{4}$$.

Combining these results gives $$S = 2 \cdot \frac{e}{4} + 3 \cdot \frac{e - e^{-1}}{4} + 4\left(\frac{e + e^{-1}}{2} - 1\right)$$ which simplifies to $$\frac{e}{2} + \frac{3e - 3e^{-1}}{4} + 2e + 2e^{-1} - 4$$, and further to $$\frac{2e + 3e - 3e^{-1} + 8e + 8e^{-1}}{4} - 4 = \frac{13e + 5e^{-1}}{4} - 4 = \frac{13e}{4} + \frac{5}{4e} - 4$$.

The correct answer is Option B: $$\frac{13e}{4} + \frac{5}{4e} - 4$$.