The sum `sum_(n=1)^oo(2n^(2)+3n+4)/((2n)!)` is equal to

Options

A. `(11e)/2+7/(2e)-4`
B. `(11e)/2+7/(2e)`
C. `(13e)/4+5/(4e)-4`
D. `(13e)/4+5/(4e)`

Answer

To solve the problem, we need to evaluate the infinite sum:

\[ S = \sum_{n=1}^{\infty} \frac{2n^2 + 3n + 4}{(2n)!} \]

### Step 1: Break down the sum

We can separate the sum into three distinct parts:

\[ S = \sum_{n=1}^{\infty} \frac{2n^2}{(2n)!} + \sum_{n=1}^{\infty} \frac{3n}{(2n)!} + \sum_{n=1}^{\infty} \frac{4}{(2n)!} \]

### Step 2: Evaluate each part

#### Part 1: Evaluate \(\sum_{n=1}^{\infty} \frac{2n^2}{(2n)!}\)

Using the identity \(n^2 = n(n-1) + n\):

\[ \sum_{n=1}^{\infty} \frac{2n^2}{(2n)!} = 2\sum_{n=1}^{\infty} \frac{n(n-1)}{(2n)!} + 2\sum_{n=1}^{\infty} \frac{n}{(2n)!} \]

The first sum can be rewritten using the factorial property:

\[ \sum_{n=1}^{\infty} \frac{n(n-1)}{(2n)!} = \sum_{n=1}^{\infty} \frac{1}{(2n-2)!} = \frac{1}{2}\sum_{m=0}^{\infty} \frac{1}{m!} = \frac{1}{2} e \]

The second sum:

\[ \sum_{n=1}^{\infty} \frac{n}{(2n)!} = \frac{1}{2}\sum_{m=0}^{\infty} \frac{1}{(2m-1)!} = \frac{1}{2} \cdot e \]

Combining these gives:

\[ \sum_{n=1}^{\infty} \frac{2n^2}{(2n)!} = 2 \left( \frac{1}{2} e + \frac{1}{2} e \right) = 2e \]

#### Part 2: Evaluate \(\sum_{n=1}^{\infty} \frac{3n}{(2n)!}\)

Using the same approach as above:

\[ \sum_{n=1}^{\infty} \frac{3n}{(2n)!} = 3 \cdot \frac{1}{2} e = \frac{3}{2} e \]

#### Part 3: Evaluate \(\sum_{n=1}^{\infty} \frac{4}{(2n)!}\)

This sum can be evaluated directly:

\[ \sum_{n=1}^{\infty} \frac{4}{(2n)!} = 4 \cdot \frac{1}{2} e = 2e \]

### Step 3: Combine all parts

Now we can combine all the parts:

\[ S = 2e + \frac{3}{2} e + 2e = 6.5 e = \frac{13}{2} e \]

### Final Result

Thus, the sum \(S\) is equal to:

\[ \frac{13}{2} e \]