Question

${\alpha }_r,{\beta }_r({\alpha }_r<{\beta }_r)$ are the roots of $x^2-r^2(r+1)x+r^5=0$. Find the value of $\sum _{r=1}^n(3{\alpha }_r+2{\beta }_r):$

Answer 1

Given equation $x^2-r^2(r+1)x+r^5=0$

${\alpha }_r,{\beta }_r$ are the roots of the equation

$\Rightarrow {\alpha }_r^2-r^2(r+1){\alpha }_r+r^5=0$

$\Rightarrow {\alpha }_r=\frac{r^2(r+1)\pm \sqrt{r^4(r+1{)}^2-4r^5}}{2}$

${\alpha }_r=\frac{r^3+r^2+r^2\sqrt{r^2+2r+1-4r}}{2}$

${\alpha }_r=\frac{r^3+r^2+r^2\sqrt{r^2-2r+1}}{2}$

${\alpha }_r=\frac{r^3+r^2+r^2(r-1)}{2}$

${\alpha }_r=\frac{r^3+r^2+r^3(r-1)}{2},{\alpha }_r=\frac{r^3+r^2-r^2(r-1)}{2}$

${\alpha }_r=\frac{r^3+r^2+r^3-r^2}{2},{\alpha }_r=\frac{r^3+r^2-r^3+r^2}{2}$

${\alpha }_r=r^3,{\alpha }_r=r^2$

Similarly ,

${\beta }_r=r^3,{\beta }_r=r^2$

${\alpha }_r<{\beta }_r$ and $1\le r\le n$

$\Rightarrow {\alpha }_r=r^2,{\beta }_r=r^3$

$\sum _{r=1}^n(3\alpha r+2\beta r):-\sum _{r=1}^{n}3r^2+2r^3$

$:-3\sum _{r=1}^n{r}^2+2\sum _{r=1}^n{r}^3$

$:-\frac{3n(n+1)(2n+1)}{6}+2{\left(\frac{n(n+1)}{2}\right)}^2$

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

$:-\frac{n(n+1)}{2}[2n+1+n^2+n]$

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