Sum of the series 1+(3^(2))/(2!)+(3^(4))...

Sum of the series `1+(3^(2))/(2!)+(3^(4))/(4!)+(3^(6))/(6!)+...."to "oo` is

`e^(-3)`

`e^(3)`

`(1)/(2)(e^(3)-e^(-3))`

`(1)/(2)(e^(3)+e^(-3))`

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The correct Answer is:

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(1)/(2!)-(1)/(3!)+(1)/(4!)-(1)/(5!)+....=
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-1+(2^(2))/(2!)-(2^(3))/(3!)+(2^(4))/(4!)-....=
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Sum of the series 1+(3^(2))/(2!)+(3^(4))/(4!)+(3^(6))/(6!)+...."to "oo...
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underset(n=0)overset(oo)Sigma(1)/((n+1)!)=
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underset(n=1)overset(oo)Sigma(1)/((2n-1)!)=
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underset(n=1)overset(oo)Sigma(x^(n))/((2n-1)!)=
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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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