1. For the system to have infinitely many solutions, the three equations must be consistent and dependent.

The equations are: $x+y+z=1$ $x+2y+4z=m$ $x+4y+10z=m^2$

So the augmented matrix must have rank less than 3, i.e. the third row should be a linear combination of the first two, and the constants must satisfy the same relation.

2. First check dependence of the coefficient rows:

Let $R_1=(1,1,1),\quad R_2=(1,2,4),\quad R_3=(1,4,10)$ Assume $R_3=aR_1+bR_2$ Then comparing coordinates: $a+b=1$ $a+2b=4$ $a+4b=10$

From the first two equations: $(a+2b)-(a+b)=4-1 \Rightarrow b=3$ Then $a=1-b=1-3=-2$ Check third: $a+4b=-2+12=10$ which is true.

Hence, $R_3=-2R_1+3R_2$ So for consistency with infinitely many solutions, the constants must also satisfy: $m^2=-2(1)+3(m)$ That is, $m^2=3m-2$ $m^2-3m+2=0$ $(m-1)(m-2)=0$

Thus, $\alpha=1,\quad \beta=2$ (up to order).

3. Now compute $\sum_{n=1}^{10}(n^{\alpha}+n^{\beta})=\sum_{n=1}^{10}(n+n^2)$

Using standard formulas: $\sum_{n=1}^{10} n=\frac{10\cdot 11}{2}=55$ $\sum_{n=1}^{10} n^2=\frac{10\cdot 11\cdot 21}{6}=385$

Therefore, $\sum_{n=1}^{10}(n+n^2)=55+385=440$

4. Hence the correct option is: $\boxed{440}$ which is Option D.