[tex]a_2=a_1+5=a_1+5+3\cdot0\\
a_3=a_2+8=a_2+5+3\cdot1\\
a_4=a_3+11=a_3+5+3\cdot2\\
a_5=a_4+14=a_4+5+3\cdot3\\
......................................................\\ [/tex]
[tex]\\
Insumam~egalitatile:\\
a_2+a_3+...+a_n=\\
=a_1+a_2++a_3+....+a_{n-1}+5(n-1)+3[1+...+(n-2)]\\
a_n=2+5n-5+3\cdot \frac{(1+n-2)(n-2)}{2} \\
a_n= \frac{2(5n-3)+3(n^2-3n+2)}{2} \\
a_n=\frac{10n-6+3n^2-9n+6}{2}\\
a_n= \frac{3n^2+n}{2} \\
Inductie~matematica\\
P(n):a_n= \frac{n(3n+1)}{2} \\
P(1):a_1=2(A)\\ Presupunem~P(n)(A)~si~aratam~P(n+1):a_{n+1}= \frac{(n+1)(3n+4)}{2} \\ a_{n+1}=a_{n}+5+3\cdot(n-1)= \frac{n(3n+1)}{2} +5+3\cdot(n-1)\\ a_{n+1}= \frac{3n^2+n+10+6n-6}{2}= \frac{3n^2+7n+4}{2} =\frac{(n+1)(3n+4)}{2} [/tex]
