Question

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

Answer 1

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

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

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

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

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

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

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

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

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

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

Similarly,

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

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

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

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

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

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

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

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

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