FOr ` xin r, x != -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 -

Options

A. `(2016!)/(16!)`
B. `(2017!)/(2000!)`
C. `(2017!)/(17!2000!)`
D. `(1016!)/(17!1999!)`

Answer

C

Explanation

We have,
`(1 + x)^(2016) + x(1 + x)^(2015) + x^(2) (1 + x)^(2014) + ..+ x^(2016) = sum_(i=0)^(2016) a_(i) x^(i)`
`rArr (1 +x)^(2016) {(((x)/(1|x))^(2017) -1)/((x)/(1+x) -1)}= sum_(i=1)^(2016) a_(1) x^(j) `
`rArr {(1 + x)^(2017) - x ^(2017)} = sum_(i=1)^(2016) a_(i) x^(j)`
`rArr a_(17) = ` Coefficient of `x^(17)` in ` {( 1 + x)^(2017) - x^(2017)}`
`rArr a_(17)` = Coefficient of `x^(17)` in `(1 + x)^(2017)`
`rArr a_(17) = ""^(2017)C_(17) = (2017!)/(17!2000!)`