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Mathematics
Problem
The sum
∑
n
=
1
∞
2
n
2
+
3
n
+
4
(
2
n
)
!
\sum_{n=1}^{\infty} \frac{2n^2 + 3n + 4}{(2n)!}
∑
n
=
1
∞
(
2
n
)!
2
n
2
+
3
n
+
4
is equal to: [Main Feb. 1, 2023 (II)] (a)
13
e
4
+
5
4
e
−
4
\frac{13e}{4} + \frac{5}{4e} - 4
4
13
e
+
4
e
5
−
4
(b)
13
e
4
+
5
4
e
\frac{13e}{4} + \frac{5}{4e}
4
13
e
+
4
e
5
(c)
11
e
2
+
7
2
e
\frac{11e}{2} + \frac{7}{2e}
2
11
e
+
2
e
7
(d)
11
e
2
+
7
2
e
−
4
\frac{11e}{2} + \frac{7}{2e} - 4
2
11
e
+
2
e
7
−
4
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