Question

REZOLVATI ECUATIA: x+(2x+1)+(3x+1)+...+(nx+n-1)=n · n

Answer 1

$x+(2x+1)+(3x+2)+ (4x+3) + ... +(nx+ n-1) =n^2$

$(x+2x+3x+4x+ ... +nx)+1+2+3+ ...+ (n-1)=n^2$

$x(1+2+3+4 + ....+n) =\dfrac{(n-1)n}{2} =n^2$

$x\cdot\dfrac{n(n+1)}{2} +\dfrac{(n-1)n}{2}=n^2$

[tex]x\cdot n(n+1) +(n-1)n=2n^2\Longrightarrow x\cdot n(n+1) = 2n^2-n^2+n \Longrightarrow x\cdot n(n+1) = n^2+n \Longrightarrow x\cdot n(n+1) = n(n+1)|_{:n(n+1)} \Longrightarrow x=1.[/tex]