If - are the roots of quadratic equation 6x2-2x-1-0 and Sn-n-n- then which of the following statements is-are true -

Given equation nbsp;6x^2 2x+1=0 has roots α, β ∴ 6α^2 2α+1=0 ⇒(6α 2+1α)=0 ⇒ nbsp;α^n(6α 2+1α)=0 ⇒ 6α^n+1 2α^n+α^n 1=0 ⋯(1) Similarly, with root β 6β^n ...

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Standard XI

Mathematics

Nature of Roots

If α,β are th...

Question

If $α, β$ are the roots of quadratic equation $6 x^{2} - 2 x + 1 = 0$ and $S_{n} = α^{n} + β^{n}$ , then which of the following statement(s) is/are true ?

6 S_{n + 1} - 2 S_{n} + S_{n - 1} = 0

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6 S_{n + 1} - 2 S_{n} + S_{n - 3} = 0

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\frac{2 S_{6} - 6 S_{7}}{S_{5}} = 1

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\frac{2 S_{6} - 6 S_{7}}{S_{5}} = 3

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Solution

The correct option is C $\frac{2 S_{6} - 6 S_{7}}{S_{5}} = 1$
Given equation $6 x^{2} - 2 x + 1 = 0$ has roots $α, β$
$∴ 6 α^{2} - 2 α + 1 = 0 \Rightarrow (6 α - 2 + \frac{1}{α}) = 0 \Rightarrow α^{n} (6 α - 2 + \frac{1}{α}) = 0 \Rightarrow 6 α^{n + 1} - 2 α^{n} + α^{n - 1} = 0 \dots (1)$

Similarly, with root $β$
$6 β^{n + 1} - 2 β^{n} + β^{n - 1} = 0 \dots (2)$

Adding equation $(1)$ and $(2)$ ,
$6 (α^{n + 1} + β^{n + 1}) - 2 (α^{n} + β^{n}) + (α^{n - 1} + β^{n - 1}) = 0 \Rightarrow 6 S_{n + 1} - 2 S_{n} + S_{n - 1} = 0 (∵ S_{n} = α^{n} + β^{n})$

Now, $S_{n - 1} = 2 S_{n} - 6 S_{n + 1}$
$\Rightarrow \frac{2 S_{n} - 6 S_{n + 1}}{S_{n - 1}} = 1 \dots (3)$
Putting the value of $n = 6$ in the equation $(3)$ , we get
$\frac{2 S_{6} - 6 S_{7}}{S_{5}} = 1$

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Similar questions

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If α,β are the roots of quadratic equation 6x2−2x+1=0 and Sn=αn+βn, then which of the following statement(s) is/are true ?

If $α, β$ are the roots of quadratic equation $6 x^{2} - 2 x + 1 = 0$ and $S_{n} = α^{n} + β^{n}$ , then which of the following statement(s) is/are true ?