[tex]a)~Avem: \\ \\ x_{n}-x_{n-1}=3 \\ \\ x_{n-1}-x_{n-2}=3 \\ \\ x_{n-2}-x_{n-3}=3 \\ \\ ........................... \\ \\ x_3-x_2=3 \\ \\ x_2-x_1=3 [/tex]
[tex]Prin~insumarea~acestor~relatii,~obtinem: \\ \\ x_n-x_1=3(n-1) \Rightarrow x_n=3(n-1)+x_1 \Leftrightarrow \boxed{x_n=3(n-1)}~.[/tex]
[tex]b)~Relatia~este~echivalenta~cu: \\ \\ 2 b_{n+1}-b_{n+1}=-2b_n+b_n \Leftrightarrow b_{n+1}=-b_{n}. \\ \\ Observam~ca~b_{n+1}=-b_n=-(-b_{n-1})=b_{n-1},~deci ~b_{n+1}=b_{n-1}.~(*)\\ \\ Din~b_{n+1}=-b_n~rezulta~b_2=-1.\\ \\ Din~b_1=1,~b_2=-1~si~din~(*)~rezulta~ca~b_{2k}=-1~si~b_{2k+1}=1 ~\forall~ \\ \\ k \in N^* \\ \\ Adica~ \boxed{ b_n=\left \{ {{1~daca~n=impar} \atop {-1~daca~n=par}} \right. } ~ .[/tex]
[tex]c)~Aici~este~mai~putin~interesant...~voi~folosi~inductia~matematica. \\ \\ Prin~calcul~obtinem: \\ \\ c_1=0=0 \cdot 1\\ \\ c_2=2=1 \cdot 2 \\ \\ c_3=6=2 \cdot 3 \\ \\ c_4=12=3 \cdot 4 \\ \\ c_5=20=4 \cdot 5 [/tex]
[tex]Pe~baza~acestor~rezultate~putem~presupune~ca~c_n=(n-1)n. \\ \\ Ramane~sa~verific~daca~este~adevarat! \\ \\ Prin~calculele~anterioare~am~verificat~prima~etapa~a~inductiei. \\ \\ In~continuare,~voi~presupune~ca~c_k=(k-1)k~(unde~k \in N^*)~si~voi~ \\ \\ demonstra~ca~c_{k+1}=k(k+1). [/tex]
[tex]Intr-adevar!~ \\ \\ c_{k+1}=1+c_k+ \sqrt{1+4 c_{k}}=1+(k-1)k+ \sqrt{1-4k+4k^2}= \\ \\ =1+(k-1)k+ \sqrt{(2k-1)^2}=1+k^2-k+2k-1=k^2+k=k(k+1), \\ \\ q.e.d. \\ \\ Deci~\boxed{c_n=(n-1)n}~ \forall~n \in N^*. [/tex]
