If `alpha, beta` are the roots of the equation `ax^(2)+bx+c=0` and `A_(n)=alpha^(n)+beta^(n)`, then `aA_(n+2)+bA_(n+1)+cA+(n)` is equal to

Options

A. 0
B. 1
C. `a+b+c`
D. `abc`

Explanation

`:' alpha +beta=-b/a` and `alpha beta=c/a`
`:.A_(n+2)=alpha^(n+2)+beta^(n+2)`
`=(alpha+beta)(alpha^(n+1)+beta^(n+1))-alphabeta^(n+1)-beta alpha^(n+1)`
`=(allpha+betaa)(alpha^(n+1)+beta^(n+1)-alpha beta(alpha^(n)+beta^(n))`
`=-b/aA_(n+1)-c/aA_(n)`
`impliesaA_(n+2)+bA_(n+1)+cA_(n)=0`