p=sum(n=0)^(oo) (x^(3n))/((3n)!) , q=sum...

`p=sum_(n=0)^(oo) (x^(3n))/((3n)!) , q=sum_(n=1)^(oo) (x^(3n-2))/((3n-2)!), r = sum_(n=1)^(oo) (x^(3n-1))/((3n-1)!)` then p + q + r =

`e^x`

`e^(2x)`

`e^(3x)`

`e^(2)`

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The correct Answer is:

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p=sum(n=0)^(oo) (x^(3n))/((3n)!) , q=sum(n=1)^(oo) (x^(3n-2))/((3n-2)!...
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