alpha_r and beta_r (alpha_r

Question

(A) The given quadratic equation is $x^2 - r^2(r + 1)x + r^5 = 0$.Let the roots be $\alpha_r$ and $\beta_r$. From the properties of roots,we have $\al

Vedclass · Accepted Answer

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

Vedclass · Answer

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

$\alpha_r$ and $\beta_r$ $(\alpha_r < \beta_r)$ are the roots of $x^2 - r^2(r + 1)x + r^5 = 0$. The value of $\sum_{r=1}^n (3\alpha_r + 2\beta_r)$ is:

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