Question

63+2•63+3•63.......+48•63=k la putere a 3a . k =?

Answer 1

     
[tex]\displaystyle 63+2 \cdot 63+3 \cdot 63+...+48 \cdot 63=k^3 \\ 63(1+2+3+ ... + 40) = k^3 \\ \\ \frac{63 \cdot 48 \cdot 49 }{2} = k^3 \\ \\ 63 \cdot 24 \cdot 49 = k^3 \\ (3^2 \times 7) \times (2^3 \times 3) \times (7^2) = k^3 \\ 2^3 \times 3^3 \times 7^3 = k^3 \\ (2 \times 3 \times 7)^3 = k^3 \\ \Longrightarrow ~~~ k = 2 \times 3 \times 7 = \boxed{42}[/tex]