If `(1+px+x^(2))^(n)=1+a_(1)x+a_(2)x^(2)+…+a_(2n)x^(2n)`.
The remainder obtained when `a_(1)+5a_(2)+9a_(3)+13a_(4)+…+(8n-3)a_(2n)` is divided by `(p+2)` is (a) 1 (b) 2 (c) 3 (d) 0

Options

A. `1`
B. `2`
C. `3`
D. `0`

Answer

C

Explanation

`(c )` `a_(1)+5a_(2)+9a_(3)+…+(8n-3)a_(2n)=sum_(r=1)^(2n)(4r-3)a_(r )`
`=4sum_(r=1)^(2n)ra_(r )-3sum_(r=1)^(2n)a_(r )`
`(1+px+x^(2))^(n)=1+a_(1)x+a_(2)x^(2)+….+a_(2n)X^(2n)`
so, `sum_(r=1)^(2n)a_(r )=(p+2)^(n)-1`
Differentiating the expansion and substituting `x=1`
`sum_(r=1)^(2n)rar_(r)=n(p+2)^(n)`
`:.sum_(r=1)^(2n)(4r-3)a_(r )=4n(p+2)^(n)-3((p+2)^(n)-1)`
`=(4n-3)(p+2)^(n)+3`