Let alpha and beta be the roots of 5x^2 - 3x | vedclass

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(D) Given the quadratic equation $5x^2 - 3x - 1 = 0$,the sum of roots is $\alpha + \beta = \frac{3}{5}$ and the product of roots is $\alpha\beta = -\f

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(D) Given the quadratic equation $5x^2 - 3x - 1 = 0$,the sum of roots is $\alpha + \beta = \frac{3}{5}$ and the product of roots is $\alpha\beta = -\frac{1}{5}$.The given series is $S = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{\alpha^n + \beta^n}{n} x^n$.This can be split into two logarithmic series: $S = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(\alpha x)^n}{n} + \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(\beta x)^n}{n}$.Using the expansion $\ln(1+t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \dots$,we get $S = \ln(1 + \alpha x) + \ln(1 + \beta x)$.$S = \ln((1 + \alpha x)(1 + \beta x)) = \ln(1 + x(\alpha + \beta) + \alpha\beta x^2)$.Substituting the values: $S = \ln(1 + x(\frac{3}{5}) - \frac{1}{5}x^2) = \ln\left(\frac{5 + 3x - x^2}{5}\right)$.

Let $\alpha$ and $\beta$ be the roots of $5x^2 - 3x - 1 = 0$. Then,the expression $\left[ (\alpha + \beta)x - \left( \frac{\alpha^2 + \beta^2}{2} \right)x^2 + \left( \frac{\alpha^3 + \beta^3}{3} \right)x^3 - \dots \right]$ is equal to:

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