Question:
Let \(\alpha\) and \(\beta\) be the roots of the equation
\(5x^2 + 6x - 2 = 0\). If \(S_n = \alpha^n + \beta^n\), \(n = 1, 2, 3, \ldots\), then :
Let \(\alpha\) and \(\beta\) be the roots of the equation
\(5x^2 + 6x - 2 = 0\). If \(S_n = \alpha^n + \beta^n\), \(n = 1, 2, 3, \ldots\), then :
\(5x^2 + 6x - 2 = 0\). If \(S_n = \alpha^n + \beta^n\), \(n = 1, 2, 3, \ldots\), then :
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Each root satisfies \(5x^2+6x-2=0\); multiply by \(x^{4}\) and add the equations for \(\alpha\) and \(\beta\).
Updated On: Nov 5, 2026
- \(5S_6 + 6S_5 = 2S_4\)
- \(5S_6 + 6S_5 + 2S_4 = 0\)
- \(6S_6 + 5S_5 + 2S_4 = 0\)
- \(6S_6 + 5S_5 = 2S_4\)
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The Correct Option is A
Solution and Explanation
Since \(\alpha\) is a root, \(5\alpha^2 + 6\alpha - 2 = 0\). Multiply by \(\alpha^4\): \(5\alpha^6 + 6\alpha^5 - 2\alpha^4 = 0\).
Similarly \(5\beta^6 + 6\beta^5 - 2\beta^4 = 0\).
Adding: \(5S_6 + 6S_5 - 2S_4 = 0\), i.e. \(5S_6 + 6S_5 = 2S_4\).
Answer: (1) \(5S_6 + 6S_5 = 2S_4\)
Similarly \(5\beta^6 + 6\beta^5 - 2\beta^4 = 0\).
Adding: \(5S_6 + 6S_5 - 2S_4 = 0\), i.e. \(5S_6 + 6S_5 = 2S_4\).
Answer: (1) \(5S_6 + 6S_5 = 2S_4\)
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