Question

Fie sirul numeric definit prin formula termenului general a(n)=2n^2+n-12 .Apartine oare numarul -2 acestui sir? dar 16?

Answer 1

$2n^{2}+n-12=-2=\ \textgreater \ 2n^2+n-12+2=0=\ \textgreater \ 2n^2+n-10=0 \\ (n-2)(2n+5)=0; n=- \frac{5}{2}; n= 2; fals \\ 2n^{2}+n-12=16=\ \textgreater \ 2n^2+n-12-16=0=\ \textgreater \ 2n^2+n-28=0=\ \textgreater \ \\ =\ \textgreater \ (2n-7)(n+4)=0; n=-4; n= \frac{7}{2}; deci fals$

-2 si 16 nu este termen acestui sir