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The sum of the series `(1)/(2!)+(1)/(4!)+(1)/(6!)+..to infty` is
A. `(e^(2)-2)/(e )`
B. `(e-1)^(2)/(2e)`
C. `(e^(2)-1)/(2e)`
D. `(e^(2)-1)/(2)`

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Answer:
We have
`e+e^(-1)=2(1+(1)/(2!)+(1)/(4!)+(1)/(6!)+..)`
`rarr (e+e^(1))/(2)-1=(1)/(2!)+(1)/(4!)+(1)/(6!)+..`
`rarr (e-1)^(2)/(2e)=(1)/(2!)+(1)/(4!)+(1)/(6!)+.`.

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