Consider `(1 + x + x^(2))^(n) = sum_(r=0)^(n) 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. `(9^(n)-2a_(2n)-1)/4`
B. `(9^(n)-2a_(2n)+1)/4`
C. `(9^(n)+2a_(2n)-1)/4`
D. `(9^(n)+2a_(2n)+1)/4`

Answer

b

Explanation

We have, `(1+x+x^(2))^(2n) = sum_(r=0)^(4n) a_(r)x^(r)` ...(i)
Replacing x by `1/x` eq. (i) We get
`(1+1/x+1/x^(2))^(2n) = sum_(r=0)^(4n) a_(r) (1/x)^(r)`
`rArr (1+x+x^(2))^(2n) = sum_(r=0) ^(4n) a_(r) x^(4n-r)` …(ii)
From Eqs. (i) and (ii) we get `sum_(r=0) ^(4n) a_(r) x^(r) = sum_(r=0)^(4n) a_(r)x^(4n-r)`
Equating the coefficient of `x^(4n-r)` on both sides, we get
`a_(4n-r) = a_(r) " for "0 le r le 4n`
Hence, `a_(r) = a_(4n-r)`
Putting `x = 1` in Eq. (i), then
`sum_(r=0)^(4n) a_(r) = 3^(2n) = 9^(n) ` ...(iii)
Putting `x = -1` in Eq (i), then `sum_(r=0)^(4n) (-1)^(r) a_(r)=1` ...(iv)
On adding Eqs. (iii) and (iv), we get
`2(a_(0) + a_(2) + a_(4) +... + a_(2n-2) + a_(2n-2) + a_(2n +...+ a_(4n) )) = 9^(n)+1`
`rArr 2[2(a_(0) + a_(2) + a_(4) + ... + a_(2n-2) ) +a_(2n))=9^(n)+1`
`[beaucse a_(r) = a_(4n-r)]`
`therefore a_(0) +a_(2) + a_(4) +... + a_(2n-2) = (9^(n)-2a_(2n)+1)/4`
`rArr sum _(r=0) ^(n-1) a_(2r) = (9^(n)- 2 a _(2n) +1)/4`