Question

Stie cineva vreun algoritm prin care sa putem sa aflam modul de evolutie al unui sir?

Ex: 2, 7 , 15 , 26, 40,....

Acest sir se poate scrie generalizat: $a_n = \frac{n(3\cdot n+1)}{2}, \quad n \geq 1$

Dar cum deduc asta?

Answer 1

[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]