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 $$\sum_{n=1}^{10} \left(n^{\alpha} + n^{\beta}\right)$$ is equal to :

For the system to have infinitely many solutions, the determinant of the coefficient matrix must be 0, and the system must be consistent.

The coefficient matrix and augmented matrix:

$$ \begin{bmatrix} 1 & 1 & 1 & | & 1 \\ 1 & 2 & 4 & | & m \\ 1 & 4 & 10 & | & m^2 \end{bmatrix} $$

$$R_2 \to R_2 - R_1$$, $$R_3 \to R_3 - R_1$$:

$$ \begin{bmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 1 & 3 & | & m-1 \\ 0 & 3 & 9 & | & m^2-1 \end{bmatrix} $$

$$R_3 \to R_3 - 3R_2$$:

$$ \begin{bmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 1 & 3 & | & m-1 \\ 0 & 0 & 0 & | & m^2-1-3(m-1) \end{bmatrix} $$

For infinite solutions: $$m^2 - 1 - 3m + 3 = 0$$

$$m^2 - 3m + 2 = 0$$

$$(m-1)(m-2) = 0$$

$$m = 1$$ or $$m = 2$$

So $$\alpha = 1, \beta = 2$$ (or vice versa).

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

$$= \frac{10 \times 11}{2} + \frac{10 \times 11 \times 21}{6} = 55 + 385 = 440$$

The correct answer is Option 4: 440.