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Question

Question: \[\frac{e^{2} + 1}{2e} =\]...

e2+12e=\frac{e^{2} + 1}{2e} =

A

1+22!+223!+234!+.....1 + \frac{2}{2!} + \frac{2^{2}}{3!} + \frac{2^{3}}{4!} + .....\infty

B

1+12!+14!+16!+.....1 + \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + .....\infty

C

12(1+12!+14!+....)\frac{1}{2}\left( 1 + \frac{1}{2!} + \frac{1}{4!} + ....\infty \right)

D

12(1+11!+12!+13!+....)\frac{1}{2}\left( 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ....\infty \right)

Answer

1+12!+14!+16!+.....1 + \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + .....\infty

Explanation

Solution

336\frac{3^{3}}{6}