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(D) For the system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $\Delta$ must be $0$,and the augmen

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$440$

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(D) For the system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $\Delta$ must be $0$,and the augmented determinants $\Delta_x, \Delta_y, \Delta_z$ must also be $0$.The coefficient matrix is $A = \begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & 4 \ 1 & 4 & 10 \end{bmatrix}$.$\Delta = 1(20-16) - 1(10-4) + 1(4-2) = 4 - 6 + 2 = 0$.Now,we check the condition for $\Delta_z = 0$ (or any other augmented determinant):$\Delta_z = \begin{vmatrix} 1 & 1 & 1 \ 1 & 2 & m \ 1 & 4 & m^2 \end{vmatrix} = 1(2m^2-4m) - 1(m^2-m) + 1(4-2) = 2m^2 - 4m - m^2 + m + 2 = m^2 - 3m + 2 = 0$.Solving $m^2 - 3m + 2 = 0$ gives $(m-1)(m-2) = 0$,so $m = 1$ or $m = 2$.Thus,$\alpha = 1$ and $\beta = 2$.We need to calculate $\sum_{n=1}^{10}(n^1 + n^2) = \sum_{n=1}^{10} n + \sum_{n=1}^{10} n^2$.Using the formulas $\sum_{n=1}^{k} n = \frac{k(k+1)}{2}$ and $\sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6}$ for $k=10$:$\sum_{n=1}^{10} n = \frac{10 	imes 11}{2} = 55$.$\sum_{n=1}^{10} n^2 = \frac{10 	imes 11 	imes 21}{6} = 385$.The total sum is $55 + 385 = 440$.

$List-I$	$List-II$
$(I)$ If $\frac{q}{r}=10$,then the system of linear equations has	$(P)$ $x=0, y=\frac{10}{9}, z=-\frac{1}{9}$ as a solution
$(II)$ If $\frac{p}{r} \neq 100$,then the system of linear equations has	$(Q)$ $x=\frac{10}{9}, y=-\frac{1}{9}, z=0$ as a solution
$(III)$ If $\frac{p}{q} \neq 10$,then the system of linear equations has	$(R)$ infinitely many solutions
$(IV)$ If $\frac{p}{q}=10$,then the system of linear equations has	$(S)$ no solution
	$(T)$ at least one solution

Let $\alpha, \beta (\alpha \neq \beta)$ 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 $\sum_{n=1}^{10}(n^\alpha+n^\beta)$ is equal to:

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