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

Options

A. `2^(19) (2^(19) -1)`
B. `2^(20)(2^(19)-1)`
C. `2^(19) (2^(20)-1)`
D. `2^(20) (2^(20)-19)`

Answer

c

Explanation

Given , ` (1 + x + 2x^(2))^(20) = a_(0) + a_(1) x + a_(2)x^(2) + …+ a_(40) x^(40)`
On putting x = 1 and x = - 1 respectively , we get
` a_(0) + a_(1) + a_(2) + …+ a_(40) = 4^(20) = 2^(40)` …(i)
and ` a_(0) - a_(1) + a_(2) - …+ a_(40) = 2^(20)` ....(ii)
From Eqs. (i) and (ii) , we get
` a_(0) + a_(2) + a_(4) + ...+ a_(38) + a_(40) = 2^(19) (2^(20)+1) `....(iii)
and ` a_(1) + a_(3) + a_(5) + ...+ a_(37) + a_(39)= 2^(19) (2^(20) -1) `...(iv)
Also , resplacing x by ` (1)/(x) ` in given expanssion , we get
`(1 + (1)/(x) + (2)/(x^(2)))^(20) = a_(0) + (a_(1))/(x) + (a_(2))/(x^(2) )+ ...+ (a_(38))/(x^(38)) + (a_(39))/(x^(39)) + (a_(40))/(x^(40) )`
` rArr (2 + x + x^(2))^(20) = a_(0) x^(40) + a_(1) x^(39) `
` + ...+ a_(38) x^(2) + a_(39) x + a_(40) ` ...(v)
On putting x = 0 , we get ` a_(40) = a^(20) ` ....(vi)
On differentiating both sides of Eq. (v) w.r.t x and put x = 0 ltbr we get
` a_(39) = 20 (2)^(19) ` ....(vii)