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

$\mathrm{\Delta }=\left|\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}10\end{array}\right|=1(20-16)-1(10-4)+1(4-2)$
= 4 – 6 + 2 = 0
For infinite solutions
$\begin{array}{l}{\mathrm{\Delta }}_{\mathrm{x}}={\mathrm{\Delta }}_{\mathrm{y}}={\mathrm{\Delta }}_{\mathrm{z}}=0\\ {\mathrm{m}}^2-3\mathrm{x}+2=0\\ \mathrm{m}=1,2\\ \mathrm{\alpha }=1,\mathrm{\beta }=2\\ \therefore \sum _{\mathrm{n}=1}^{10} \left({\mathrm{n}}^{\mathrm{\alpha }}+{\mathrm{n}}^{\mathrm{\beta }}\right)=\sum _{\mathrm{n}=1}^{10} {\mathrm{n}}^1+\sum _{\mathrm{n}=1}^{10} {\mathrm{n}}^2\\ =\frac{10\left(11\right)}{2}+\frac{10\left(11\right)\left(21\right)}{6}\end{array}$
= 55 + 385
= 440