Consider `(1 + x + x^(2))^(n) = sum_(r=0)^(2n) a_(r) x^(r)` , where ` a_(0), a_(1), a_(2),…, a_(2n)` are
real number and n is positive integer.
If n is even, the value of ` sum_(r=0)^(n/2-1) a_(2r) ` is

Options

A. `(3^(n) - 1 + a_(n))/(2)`
B. `(3^(n) -1 - a_(n))/(4)`
C. `(3^(n) + 1 + a_(n))/(2)`
D. `(3^(n) + 1 - 2a_(n))/(4)`

Answer

d

Explanation

On putting x = 1 and x = -1 in Eq (i) , we get
` 3^(n) = a_(0) + a_(1) + a_(2) + …+ a_(2n) ` ….(v)
` 1= a_(0) - a_(1) + a_(2) -a_(3) + …+ a_(2n) ` …(vi)
On adding and subtracting Eqs (v) and (vi) , we get
` (3^(n) + 1)/(2)= (a_(0) + a_(2) + ...+ a_(2n) ) ` ...(vii)
`(3^(n) -1)/(2) = (a_(1) + a_(3) + a_(5) + ...+ a_(2n-1))` ....(viii)
Also , ` a_(r) = a_(2n-r)`
Put r = 0, 2,4,6,.., n-1
` a_(0) = a_(2n), a_(2n -2), a_(4) = a_(2n-4) ,...`
` a_(n-1) = a_(n+1)`
From Eq. (vii) , we get
` (3^(n) +1)/(2) = 2 (a_(0) + a_(2) + ...+ a_(n-2)) + a_(n)`
` (3^(n) +1- 2a_(n))/(4) = a_(0) + a_(2) + ...+ a_(n-1)`