If α and βare the roots of equation x2-3x+1=0 and an=αn+βn,n∈N then the value of a7+a5a6 is

It is given that

α

and

β

are the roots of equation

x^{2} - 3 x + 1 = 0

and

a_{n} = α^{n} + β^{n}, n \in N .

The standard quadratic equation is

a x^{2} + bx + c = 0 .

We know that the sum of roots of a quadratic equation is =

- \frac{b}{a}

And the product of the roots of a quadratic equation is

= \frac{c}{a} .

By using the above relations, we can find the value of k.
We can write that,

αβ = 1, α + β = 3

.
Now, for

a_{n} = α^{n} + β^{n}

= \frac{a_{7} + a_{5}}{a_{6}}

= \frac{α^{7} + β^{7} + α^{5} + β^{5}}{α^{6} + β^{6}}

= \frac{α}{1 + {(\frac{β}{β})}^{6}} + \frac{β}{(1 + {(\frac{β}{α})}^{6})} \frac{1}{α (1 {(\frac{β}{α})}^{6})} + \frac{1}{β (1 + {(\frac{α}{β})}^{6})}

We get the condition,

\frac{α}{β} = \frac{β}{α} = 1

= \frac{α}{2} + \frac{β}{2} + \frac{1}{2 α} + \frac{1}{2 β}

= \frac{3}{2} + \frac{α + β}{2 αβ}

= \frac{3}{2} + \frac{3}{2}

= 3

Hence, the correct option is 3.

If $α$ and $β$ are the roots of equation $x^{2} - 3 x + 1 = 0$ and $a_{n} = α^{n} + β^{n}, n \in N$ then the value of $\frac{a_{7} + a_{5}}{a_{6}}$ is