Let a, b, and c be three real numbers satifying `[(a, b, c)]`
`[(1,9,7),(8,2,7),(7,3,7)]=[(0,0,0)]`
Let `b=6`, with a and c satisfying (E). If `alpha` and `beta` are the roots of the quadratic equation `ax^(2)+bx+c=0`, then `sum_(n=0)^(oo) (1/alpha+1/beta)^(n)` is

Options

A. 6
B. 3
C. `6/7`
D. `infty`

Answer

B

Explanation

`because b= 6, ` with a and c satisgying ( E)
`therefore a + 48 + 7 c = 0, 9a + 12 + 3c= 0, a + 6 +c=0`
we get `a = 1, c -7`
Given, `alpha ,beta` are the roots of `ax^(2) + bx+c=0`
`therefore alpha + beta = -b/a = -6, `
`alpha beta = c/a = -7`
Now, `1/alpha + 1/beta=(alpha + beta)/(alpha beta) = (-6)/(-7) = 6/7`
`therefore sum _(n=0) ^(infty) (1/alpha +1/beta)^(n) = sum _(n=0)^(infty)(6/7)^(n) `
`= 1+ ( 6/7) + (6/7) ^(2) +...infty`
`1/(1-6//7)=7`