If `(1 + x+ 2x^(2))^(20) = a_(0) + a_(1) x + a_(2) x^(2) + …+ a_(40) x^(40)` .
The value of ` a_(0) + a_(2) + a_(4) + …+ a_(38)` is

Explanation

` ( 1 + x - 2x^(2))^(20) = sum_(r=0)^(40) a_(r) x^(r) ` …(i)
Putting x = 1 , we get ` 0= sum_(r=0)^(a0) a^(r)`
or `a_(0)+a_(1)+a_(2)+a_(3)+ a_(4)+ a_(5)+...+ a_(39) +a_(40) = 0 ` ...(ii)
Putting x = - 1 in Eq . (i) , we get
`(-2)^(20) = sum_(r=0)^(40) (-1)^(r) a^(r)`
`a_(0)-a_(1)-a_(2)-a_(3)- a_(4)- a_(5)+...- a_(39) +a_(40) = 2^(20) ` ...(iii)
On subtracting . Eq (iii) from Eq. (ii) , we get
`2[a_(1) + a_(3) + a_(5) + ...+ a_(39) ] = 2^(20)`
or ` a_(1) + a_(3) + a_(5) + a_(39) = 0 2^(19)`
Corollary On adding Eqs. (ii) and (iii) and then dividing by
2 we get ` a_(0) + a_(2) + a_(4) + ... + a_(40) + 2^(19)` .