If alpha_r and beta_r (where alpha_r

Question

(A) The given quadratic equation is $x^2 - r^2(r + 1)x + r^5 = 0$. By Vieta's formulas,the sum of roots $\alpha_r + \beta_r = r^2(r + 1) = r^3 + r^2$

Vedclass · Accepted Answer

$\frac{1}{2}n(n + 1)(n^2 + 3n + 1)$

Vedclass · Answer

(A) The given quadratic equation is $x^2 - r^2(r + 1)x + r^5 = 0$. By Vieta's formulas,the sum of roots $\alpha_r + \beta_r = r^2(r + 1) = r^3 + r^2$ and the product of roots $\alpha_r \beta_r = r^5$. Since $\alpha_r \beta_r = r^2 \cdot r^3 = r^5$ and $\alpha_r + \beta_r = r^2 + r^3$,the roots are $\alpha_r = r^2$ and $\beta_r = r^3$ (given $\alpha_r  1$). We need to calculate $S = \sum_{r=1}^{n} (3\alpha_r + 2\beta_r) = \sum_{r=1}^{n} (3r^2 + 2r^3)$. Using the standard summation formulas $\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{r=1}^{n} r^3 = \frac{n^2(n+1)^2}{4}$: $S = 3 \cdot \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n^2(n+1)^2}{4} = \frac{n(n+1)(2n+1)}{2} + \frac{n^2(n+1)^2}{2}$. Factoring out $\frac{n(n+1)}{2}$: $S = \frac{n(n+1)}{2} [ (2n+1) + n(n+1) ] = \frac{n(n+1)}{2} [ 2n + 1 + n^2 + n ] = \frac{n(n+1)(n^2 + 3n + 1)}{2}$.

If $\alpha_r$ and $\beta_r$ (where $\alpha_r < \beta_r$) are the roots of the quadratic equation $x^2 - r^2(r + 1)x + r^5 = 0$,then find the value of $\sum_{r=1}^{n} (3\alpha_r + 2\beta_r)$.

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