Let alpha and beta be the roots of the equati | vedclass

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(A) Since $\alpha$ and $\beta$ are the roots of the equation $5x^{2} + 6x - 2 = 0$,they satisfy the equation.Therefore,$5\alpha^{2} + 6\alpha - 2 = 0$

Vedclass · Accepted Answer

$5S_{6} + 6S_{5} = 2S_{4}$

Vedclass · Answer

(A) Since $\alpha$ and $\beta$ are the roots of the equation $5x^{2} + 6x - 2 = 0$,they satisfy the equation.Therefore,$5\alpha^{2} + 6\alpha - 2 = 0$ and $5\beta^{2} + 6\beta - 2 = 0$.Multiplying the first equation by $\alpha^{n}$ and the second by $\beta^{n}$,we get:$5\alpha^{n+2} + 6\alpha^{n+1} - 2\alpha^{n} = 0$ ... $(1)$$5\beta^{n+2} + 6\beta^{n+1} - 2\beta^{n} = 0$ ... $(2)$Adding equations $(1)$ and $(2)$,we get:$5(\alpha^{n+2} + \beta^{n+2}) + 6(\alpha^{n+1} + \beta^{n+1}) - 2(\alpha^{n} + \beta^{n}) = 0$Given $S_{n} = \alpha^{n} + \beta^{n}$,this simplifies to:$5S_{n+2} + 6S_{n+1} - 2S_{n} = 0$For $n = 4$,we substitute into the relation:$5S_{6} + 6S_{5} - 2S_{4} = 0$Rearranging the terms gives:$5S_{6} + 6S_{5} = 2S_{4}$

Let $\alpha$ and $\beta$ be the roots of the equation $5x^{2} + 6x - 2 = 0$. If $S_{n} = \alpha^{n} + \beta^{n}$ for $n = 1, 2, 3, \ldots$,then:

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