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For x in R, x ne -1, if (1+x)^(2016)...

For `x in R, x ne -1`, if
`(1+x)^(2016)+x(1+x)^(2015)+x^(2)(1+x)^(2014)+……+x^(2016) = sum_(i=0)^(2016)a_(i)*x^(i)`, then `a_(17)` is equal to

A

`(2016!)/(17!1999!)`

B

`(2016!)/(16!)`

C

`(2017!)/(2000!)`

D

`(2017!)/(17!2000!)`

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