If `(1+px+x^(2))^(n)=1+a_(1)x+a_(2)x^(2)+…+a_(2n)x^(2n)`.
Which of the following is true for `1 lt r lt 2n`

Options

A. (a) `(np+pr)a_(r )=(r+1)a_(r+1)+(r-1)a_(r-1)`
B. (b) `(np-pr)a_(r )=(r+1)a_(r+1)+(r-1-2n)a_(r-1)`
C. (c) `(np-pr)a_(r )=(r+1)a_(r+1)+(r-1-n)a_(r-1)`
D. (d) `(2np+pr)a_(r )=(r+1+n)a_(r+1)+(r+1-n)a_(r-1)`

Answer

B

Explanation

`(b)` Differntiating the expansion we have
`n(p+2x)(1+px+x^(2))^(n-1)`
`=a_(1)+2a_(2)x+3a_(3)x^(2)+….+2na_(2n)x^(2n-1)`
Multiplying by `(1+px+x^(2))`
`n(p+2x)(1+a_(1)x+a_(2)x^(2)+….)`
`=(1+px+x^(2))(a_(1)+2a_(2)x+3a_(3)x^(2)+...+2na_(2n)x^(2n-1))`
Comparing coefficient of `x^(r )` both side.
`n[pa_(r )+2a_(r-1)]=(r+1)a_(r+1)+pra_(r )+(r-1)a_(r-1)`
`:.(np-pr)a_(r )=(r+1)a_(r+1)+(r-1-2n)a_(r-1)`