If `(1-x-x^(2))^(20) = sum_(r=0)^(40)a_(r).x^(r )`, then value of `a_(1) + 3a_(3) + 5a_(5) + "….." + 39a_(39)` is

Answer

40

Explanation

`(1-x-x^(2))^(20) = underset(r=0)overset(40)suma_(r).x^(r)`
Differentiating both sides w.r.t.x, we get
`20 (1-x-x^(2))^(19)(-1-2x) = underset(r=0)overset(40)suma_(r).x^(r-1)"........."(1)`
Putting `x = 1` into `(1)`, we get
`20(1-4)^(19)(-1-2) = underset(r=0)overset(40)sumra_(r)`
`rArr 60 = a_(1) + 2a_(2) + 3a_(3) + "...." + 40a_(40) "......."(2)`
Putting `x = - 1` into `(1)`, we get
`20(1+1-1)^(19)(-1+2) = underset(r=0)overset(40)sumra_(r)(-1)^(r-1)`
`rArr 20 = a_(1) - 2a_(2) + 3a_(3) + "......" - 40 a_(40) "......."(3)`
Adding `(2)` and `(3)`, we get
`80 = 2a_() + 6a_(3) + "....." + 78a_(39)`
`:. a_(1) + 3a_(3) + "....." + 39a_(39) = 40`