Question

brainly challange solve it.......

Answer 1

Hey friend, Harish here.

Here is your answer:

Given that:

→ $\alpha , \beta , \gamma \ or \ the \ roots\ of \ x^3 - 2x^2 -1=0.$

→ $T_n = ( \alpha )^n + (\beta )^n + (\gamma)^n$

To find:

$The\ value\ of\ \frac{T_{11} - T_8 }{T_{10}}$

Solution:

As α, β , γ are the roots ,  these will be satisfying the equation when substituted in place of x.

The we get, 

⇒ $\alpha ^3 - 2 \alpha ^2 -1 = 0$

⇒ $\alpha ^3 -1 = 2 \alpha ^2$  - (i)

⇒ $\beta ^3 - 2 \beta ^2 - 1 =0$

⇒ $\beta ^3 - 1 = 2 \beta ^2$  - (ii)

⇒ $\gamma ^3 - 2\gamma ^2 -1 =0$

⇒ $\gamma ^3 - 1 = 2 \gamma ^2$ - (iii)

Now,

 ⇒ $\frac{T_{11} - T_8 }{T_{10}} = \frac{[ \alpha ^{11} + \beta ^{11} + \gamma^{11} - \alpha ^{11} - \beta ^{11} - \gamma^{8}]}{(\alpha ^{10} + \beta ^{10} + \gamma^{10})}$

⇒ $\frac{\alpha ^{11} - \alpha ^8 + \beta ^{11} - \beta ^8 + \gamma^{11}- \gamma ^8}{\alpha ^{10} + \beta ^{10} + \gamma^{10}}$

⇒ $\frac{\alpha ^8(\alpha^3 -1) + \beta ^8 (\beta ^3 - 1) + \gamma ^8 (\gamma ^3 - 1)}{\alpha ^{10} + \beta ^{10} + \gamma^{10}}$

Now substituting the values from (i) , (ii) , (iii) we get,

⇒ $\frac{2\alpha ^{10} + 2\beta ^{10} + 2\gamma^{10}}{\alpha ^{10} + \beta ^{10} + \gamma^{10}} = \frac{2\times (\alpha ^{10} + \beta ^{10} + \gamma^{10})}{(\alpha ^{10} + \beta ^{10} + \gamma^{10})}$

⇒ $\frac{T_{11} - T_8 }{T_{10}} = 2.$
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Hope my  answer is helpful to you.