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 numbers and n is a positive integer.
The value of `a_(2)` is

Options

A. `""^(4n+1)C_(2)`
B. `""^(3n+1)C_(2)`
C. `""^(2n+1)C_(2)`
D. `""^(n+1)C_(2)`

Answer

c

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)
`because a_(2) = "Coefficient fo " x^(2) " in" (1 + x + x^(2))^(2n)`
`therefore (1+ x + x^(2)) ^(2n) = sum_(alpha + beta +gamma=2n ) (2n!)/(alpha !beta ! gamma!) (1)^(alpha) (x)^(beta) (x^(2))^(gamma)`
`= sum_(alpha + beta +gamma=2n ) (2n!)/(alpha !beta ! gamma!) x^(beta + 2 gamma )`
For `a_(2), beta+ 2gamma = 2 ` Possible values of `alpha, beta, gamma `are (2n-2,2,0) and (2n- 1, 0, 1).
`therefore a_(2) = (2n!)/((2n-2)! 2!0!)+ (2n!)/((2n-1)!0!1!)`
`= ""^(2n)C _(2)+""^(2n) C_(1) =""^(2n+1) C_(2)`