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If a(n) = ( log(e ) 3)^(n) sum(k!(n-k)!)...

If `a_(n) = ( log_(e ) 3)^(n) sum_(k!(n-k)!)^(k^(2)) ` then `a_(1) + a_(2) +a_(3) +"...."oo` is equal to

A

`3 log_(e ) 9`

B

`9 log_(e ) 3`

C

`9 log_(e ) 3 ( log_(e )3+1)`

D

`( log_(e ) 9)^(2)`

Text Solution

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The correct Answer is:
C
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