Let , () be the values of m for which the equations x+y+z=1, x+2y+4z=m, x+4y+10z=m^2 have …

Question

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

Accepted Answer

$\Delta=\begin{vmatrix}1&1&1\1&2&4\1&4&10\end{vmatrix}=1(20-16)-1(10-4)+1(4-2)=4-6+2=0$ for all $m$, so infinitely many solutions requires the remaining determinant conditions, which reduce to
$$m^2-3m+2=0\Rightarrow m=1,2.$$

So $\alpha=1,\ \beta=2$.

$$\sum_{n=1}^{10}(n^1+n^2)=\sum_{n=1}^{10}n+\sum_{n=1}^{10}n^2=\frac{10\cdot11}2+\frac{10\cdot11\cdot21}6=55+385=440.$$

Solutions · 1