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 odd , the value of ` sum_(r-1)^(2) a_(2r -1)` is

Options

A. `(9^(n)-1)/2`
B. `(9^(n)-1)/4`
C. `(9^(n)+1)/2`
D. `(9^(n)+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 subtracting Eq. (iv) from Eq. (iii), we get
`2 (a_(1) + a_(3) + a_(5) +... + a_(2n-1) + a_(4n-1))= 9^(n-1)`
`rArr 2[2(a_(1) + a_(3) +a_(5) + ... + a_(2n-1)] = 9^(n) - 1 [because a_(r) = a_(4n-r)]`
`therefore a_(1) + a_(3) + a_(5) + ... + a_(2n-1) = (9^(n-1))/4 `
` rArr sum_(r=1)^(n) a_(2r-1) = (9^(n)-1)/4`