Question

determinati numarul natural x pentru care numerele x+1,x^2+1 si x^3-1 sunt termeni consecutivi ai unei progresii aritmetice

Answer 1

$x+1,~x^2+1~si~x^3-1~sunt~termeni~consecutivi~ai~unei~progresii~aritmetice.\\ \\a_{n-1}=x+1\\ \\ a_{n}=x^2+1\\ \\ a_{n+1}=x^3-1\\ \\ Termenii~a_{n-1},~a_{n},~a_{n+1}~sunt~termenii~unei~progresii~aritmetice~daca:\\ \\ \boxed{a_{n}=\frac{a_{n-1}+a_{n+1}}{2}}\\ \\ \\ \\ Deci~daca~x^2+1=\frac{(x+1)+(x^3-1)}{2}\\ \\ <=>~2(x^2+1)=(x+1)+(x^3-1)\\ \\ <=>~2x^2+2=x+x^3\\ \\ <=>~x^3-2x^2+x-2=0\\ \\ <=>~x^2(x-2)+(x-2)=0\\ \\<=>~ (x-2)(x^2+1)=0\\ \\ \left \{ {{x-2=0 ~\Rightarrow ~x=2 ~\in ~N} \atop {x^2+1=0,~fals~deoarece~x^2\geq0~|+1 \Rightarrow x^2+1\geq1 }} \right\\ \\ \\ \\ \boxed{S={2}}$