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

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

Decompose the numerator using factorial identities.

We write $$2n^2 + 3n + 4 = n(2n-1) + 4n + 4$$.

Evaluate each part separately.

Part A: $$\sum_{n=1}^{\infty} \frac{n(2n-1)}{(2n)!}$$

Since $$(2n)! = (2n)(2n-1)(2n-2)!$$, we get:

$$\frac{n(2n-1)}{(2n)!} = \frac{n(2n-1)}{(2n)(2n-1)(2n-2)!} = \frac{1}{2(2n-2)!}$$

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

Part B: $$\sum_{n=1}^{\infty} \frac{4n}{(2n)!}$$

$$\frac{4n}{(2n)!} = \frac{4n}{(2n)(2n-1)!} = \frac{2}{(2n-1)!}$$

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

Part C: $$\sum_{n=1}^{\infty} \frac{4}{(2n)!}$$

$$4\sum_{n=1}^{\infty} \frac{1}{(2n)!} = 4\left(\cosh(1) - 1\right) = 4\left(\frac{e + e^{-1}}{2} - 1\right) = 2(e + e^{-1}) - 4$$

Add all parts.

$$S = \frac{e + e^{-1}}{4} + (e - e^{-1}) + 2(e + e^{-1}) - 4$$

$$= \frac{e}{4} + \frac{1}{4e} + e - \frac{1}{e} + 2e + \frac{2}{e} - 4$$

$$= \left(\frac{1}{4} + 1 + 2\right)e + \left(\frac{1}{4} - 1 + 2\right)\frac{1}{e} - 4$$

$$= \frac{13e}{4} + \frac{5}{4e} - 4$$

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