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:
Dar cum deduc asta?
matepentrutoti
![\\ 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}](https://tex.z-dn.net/?f=%5C%5C%0AInsumam%7Eegalitatile%3A%5C%5C%0Aa_2%2Ba_3%2B...%2Ba_n%3D%5C%5C%0A%3Da_1%2Ba_2%2B%2Ba_3%2B....%2Ba_%7Bn-1%7D%2B5%28n-1%29%2B3%5B1%2B...%2B%28n-2%29%5D%5C%5C%0Aa_n%3D2%2B5n-5%2B3%5Ccdot+%5Cfrac%7B%281%2Bn-2%29%28n-2%29%7D%7B2%7D+%5C%5C%0Aa_n%3D+%5Cfrac%7B2%285n-3%29%2B3%28n%5E2-3n%2B2%29%7D%7B2%7D+%5C%5C%0Aa_n%3D%5Cfrac%7B10n-6%2B3n%5E2-9n%2B6%7D%7B2%7D%5C%5C%0Aa_n%3D+%5Cfrac%7B3n%5E2%2Bn%7D%7B2%7D+%5C%5C%0AInductie%7Ematematica%5C%5C%0A%0AP%28n%29%3Aa_n%3D+%5Cfrac%7Bn%283n%2B1%29%7D%7B2%7D+%5C%5C%0AP%281%29%3Aa_1%3D2%28A%29%5C%5C+Presupunem%7EP%28n%29%28A%29%7Esi%7Earatam%7EP%28n%2B1%29%3Aa_%7Bn%2B1%7D%3D+%5Cfrac%7B%28n%2B1%29%283n%2B4%29%7D%7B2%7D+%5C%5C+a_%7Bn%2B1%7D%3Da_%7Bn%7D%2B5%2B3%5Ccdot%28n-1%29%3D+%5Cfrac%7Bn%283n%2B1%29%7D%7B2%7D+%2B5%2B3%5Ccdot%28n-1%29%5C%5C+a_%7Bn%2B1%7D%3D+%5Cfrac%7B3n%5E2%2Bn%2B10%2B6n-6%7D%7B2%7D%3D+%5Cfrac%7B3n%5E2%2B7n%2B4%7D%7B2%7D+%3D%5Cfrac%7B%28n%2B1%29%283n%2B4%29%7D%7B2%7D%C2%A0 "\\ 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}")
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