Let $\alpha, \beta(\alpha
eq \beta)$ be the values of m , for which the equations $x+y+z=1 ; x+2 y+4 z=\mathrm{m}$ and $x+4 y+10 z=m^2$ have infinitely many solutions. Then the value of $\sum_{n=1}^{10}\left(n^\alpha+n^\beta ight)$ is equal to :

Question

Let $\alpha, \beta(\alpha 
eq \beta)$ be the values of m , for which the equations $x+y+z=1 ; x+2 y+4 z=\mathrm{m}$ and $x+4 y+10 z=m^2$ have infinitely many solutions. Then the value of $\sum_{n=1}^{10}\left(n^\alpha+n^\betaight)$ is equal to :

Aicharya · Accepted Answer

$\begin{aligned}
\Delta & =\left|\begin{array}{llc}
1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10
\end{array}\right|=1(20-16)-1(10-4)+1(4-2) \\ & =4-6+2=0
\end{aligned}$
For infinite solutions
$\begin{aligned}
& \Delta_x=\Delta_y=\Delta_z=0 \\ & m^2-3 x+2=0 \\ & m=1,2 \\ & \alpha=1, \beta=2 \\ & \therefore \sum_{n=1}^{10}\left(n^\alpha+n^\beta\right)=\sum_{n=1}^{10} n^1+\sum_{n=1}^{10} n^2 \\ &= \frac{10(11)}{2}+\frac{10(11)(21)}{6} \\ &= 55+385 \\ &= 440
\end{aligned}$

Let $\alpha, \beta(\alpha \neq \beta)$ be the values of m , for which the equations $x+y+z=1 ; x+2 y+4…

Solution