Question

If `(1+x-2x^2)^(20)=a_0a_1x=a_2x^2+a_3x^3++a_(40)x^(40),` then find the value of `a_1+a_3+a_5++a_(39)dot`

Answer 1

`(1+x-2x^(2))^(20) = a_(0) + a_(1)x+a_(2)x^(2) + "….." +a_(40)x^(40)`
Putting `x = 1`, we get
`a_(0) + a_(1) + a_(2) + a_(3) +"……."+a_(40) = 0" "(1)`
Putting, `x = - 1`, we get
`a_(0) - a_(1) + a_(2) - a_(3) + "……" - a_(39) + a_(40) = 2^(20) " "(2)`
Substracting (2) from (1), we get
`2[a_(1)+a_(3)+"......"+a_(39)]=-2^(20)`
or `a_(1)+a_(3)+"....."+a_(39)=-2^(19)`