(2)/(1!)+(4)/(3!)+(6)/(5!)+....=...

`(2)/(1!)+(4)/(3!)+(6)/(5!)+....=`

`e^(2)`

`2e`

`1//e`

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The correct Answer is:

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The sum of series (1)/(2!)+(1)/(4!)+(1)/(6!)+... is
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1+(1)/(3!)+(1)/(5!)+....=
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(2)/(1!)+(4)/(3!)+(6)/(5!)+....=
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(3)/(2!)+(5)/(4!)+(7)/(6!)+(9)/(8!)+....=
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2[1+((log(e)a)^(2))/(2!)+((log(e)a)^(4))/(4!)+....]=
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1+(2)/(3!)+(3)/(5!)+(4)/(7!)+....=
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(2)/(3!)+(4)/(5!)+(6)/(7!)+....=
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(1)/(3!)+(2)/(5!)+(3)/(7!)+....=
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1+(2)/(1!)+(3)/(2!)+(4)/(3!)+....=
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1+(3)/(1!)+(5)/(2!)+(7)/(3!)+....=
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(3)/(2!)+(5)/(3!)+(7)/(4!)+(9)/(5!)+....=
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1+(4)/(2!)+(7)/(3!)+(10)/(4!)+....=
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underset(n=1)overset(oo)Sigma(2n)/((2n+1)!)=
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underset(n=1)overset(oo)sum(n^(2))/((n+1)!)=
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underset(n=1)overset(oo)Sigma(2n^(2)+n+1)/(n!)=
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(1+(1)/(2!)+(1)/(4!)+(1)/(6!)+....)^(2)-(1+(1)/(3!)+(1)/(5!)+(1)/(7!)+...
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(1+(a^(2)x^(2))/(2!)+(a^(4)x^(4))/(4!)+....)^(2)-(ax+(a^(3)x^(3))/(3!)...
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(1+(2^(2))/(2!)+(2^(4))/(3!)+(2^(6))/(4!)+....)/(1+(1)/(2!)+(2)/(3!)+(...
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(1+(1)/(3!)+(1)/(5!)+....)/(1+(1)/(2!)+(1)/(4!)+....)=
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((1)/(2!)+(1)/(4!)+(1)/(6!)+....)/(1+(1)/(3!)+(1)/(5!)+....)=
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