Question

Am nevoie de rezolvarea exercitiului !!!​

Answer 1

$\displaystyle\sf\\b)~f(1)+f(2)+f(3)+...+f(n)=105,~observam~ca~1,2,3,...,n\in[1,+\infty),\\prin~urmare~functia~f(n)~va~fi~3n-6,~conform~cerintei.\\f(1)=3\cdot1-6=-3.\\f(2)=3\cdot2-6=0.\\f(3)=3\cdot3-6=3.\\.\\.\\.\\f(n)=3n-6.\\f(1)+f(2)+...+f(n)=-3+0+3+...+(3n-6)=\\6+9+12+...+(3n-6)=105 |:3 \implies\\2+3+4+...+(n-2)=35|+1 \implies 1+2+3+...+(n-2)=36 \implies\\\frac{(n-1)(n-2)}{2}=36 \implies n^2-3n+2=72 \Leftrightarrow\\n^2-7n-10n-70=0 \implies (n+7)(n-10)=0 \implies n=10,~n~poate$

$\displaystyle\sf\\fi~si~-7,~dar~problema~spune~ca~n\in\mathbb{N}.$