If `(1-x+x^2)^n=a_0+a_1x+a_2x^2+ .........+a_(2n)x^(2n),\ ` find the value of `a_0+a_2+a_4+........+a_(2n)dot`

Options

A. `sum_(r=0)^(2n) a_(r) = 3^(n)`
B. `a_(r) = a_(2n-r) - le rle 2n`
C. `a_(0) + a_(1) + a_(2) + ...+ a_(n-1) = (3^(n) - a_(n))/(2)`
D. `a_(0) + a_(1) + a_(2) + ...+ a_(n) = (3^(n))/(2)`

Answer

d

Explanation

(i) We have ,
`(1 + x + x ^(2))^(n) = a_(0)+ a_(1) + x + a_(2) x^(2) + ...+ a_(2n) x^(2n)`
Putting x = 1 on both sides, we get
`a_(0) + a_(1) x + a_(2)+...+ a_(2n) = 3^(n)`
(ii) We have,
`(1 + x + x ^(2))^(n) =sum_(r=0)^(2n)a_(r) x^(r)` ...(i)
`rArr ({:( " 1" " 1"),("1" " |"" |"),(" "x " "x^(2)):})^(n) = sum_(r=0)^(2n) {:(a_(r)),(x^(r)):}` ""[Replacing x by 1/x]
`rArr (1 + x + x^(2) = sum_(r=0)^(2n) a_(r) x^(2n - r)` ...(ii)
From (i) and (ii), we get
`sum_(r=0)^(2n) a_(r) x^(r) = sum_(r=0)^(2n) a_(r) x^(2n - r)` [From (i) and (ii)]
`rArr a_(2n -r) = a _(r) ` for 0 `le r lt 2n [{:("On equating the coefficient of "),( " "x^(2n-r)"on the sides"):}]`
(iii) We have,
`a_(r) = a_(2n - r) ` for 0` lt r le 2n`
`rArr sum_(r=0)^(n-1) a_(r) = sum_(r=0)^(n-1) a_(2n-r)`
`rArr 2(a_(0) + a_(1) +...+ a_(n-1)) + a_(n) = a_(0)+a_(1)+a_(2) + ...+ a_(2n)`
`rArr 2( a_(0) + a_(1) +...+a_(n-1) )+ a_(n) = 3^(n) "" [because sum_(r=0)^(2n) a_(r) = 3^(n) ]`
`rArr a_(0) + a_(1) + ...+ a_(n-1) = (1)/(2) (3^(n) - a_(n))`.