Question

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

• A. \(\frac{11 e }{2}+\frac{7}{2 e }-4\)
• B. \(\frac{11 e }{2}+\frac{7}{2 e }\)
• C. \(\frac{13 e }{4}+\frac{5}{4 e }-4\)
• D. \(\frac{13 e }{4}+\frac{5}{4 e }\)

Hint:

For sums involving factorials, try to decompose the terms into known series expansions like e^x or related expressions. This makes it easier to compute and simplify the results.

Answer 1

The Correct Option is C

The given sum is:
\[\sum_{n=1}^\infty \frac{2n^2 + 3n + 4}{(2n)!}.\]
Step 1: Split the Terms of the Numerator
Rewrite the numerator \(2n^2 + 3n + 4\) as:
\[2n^2 + 3n + 4 = 2n(2n - 1) + 8n + 8.\]
Thus, the sum becomes:
\[\sum_{n=1}^\infty \frac{2n^2 + 3n + 4}{(2n)!} = \frac{1}{2} \sum_{n=1}^\infty \frac{2n(2n - 1)}{(2n)!} + 2 \sum_{n=1}^\infty \frac{n}{(2n - 1)!} + 4 \sum_{n=1}^\infty \frac{1}{(2n)!}.\]
Step 2: Simplify Each Term Using Series Expansions
For the first term:
\[\sum_{n=1}^\infty \frac{2n(2n - 1)}{(2n)!} = \sum_{n=1}^\infty \frac{1}{(2n - 2)!}.\]
This is the series expansion of \(e + \frac{1}{2}\), so:
\[\frac{1}{2} \sum_{n=1}^\infty \frac{2n(2n - 1)}{(2n)!} = \frac{e + \frac{1}{2}}{2}.\]
For the second term:
\[\sum_{n=1}^\infty \frac{n}{(2n - 1)!} = e - \frac{1}{e}.\]
Thus:
\[2 \sum_{n=1}^\infty \frac{n}{(2n - 1)!} = 2 \left(e - \frac{1}{e}\right).\]
For the third term
\[\sum_{n=1}^\infty \frac{1}{(2n)!} = \frac{e + \frac{1}{2}}{2}.\]
Thus:
\[4 \sum_{n=1}^\infty \frac{1}{(2n)!} = 2 \left(e + \frac{1}{e}\right).\]
Step 3: Combine All Terms
Combine all terms:
\[\frac{1}{2} \sum_{n=1}^\infty \frac{2n(2n - 1)}{(2n)!} + 2 \sum_{n=1}^\infty \frac{n}{(2n - 1)!} + 4 \sum_{n=1}^\infty \frac{1}{(2n)!}.\]
Simplify:
\[\text{Sum} = \frac{e + \frac{1}{2}}{2} + 2 \left(e - \frac{1}{e}\right) + 2 \left(e + \frac{1}{e}\right).\]
Combine terms:
\[\text{Sum} = \frac{e}{2} + \frac{1}{4} + 2e - \frac{2}{e} + 2e + \frac{2}{e}.\]
\[\text{Sum} = 3e + \frac{1}{4}.\]
Conclusion
The sum is:
\[\frac{13e}{4} + \frac{5}{4e} - 4 \quad.\]