Categorie: Referate MatematicaReferat downloadat de: 334 ori.
Cauta referat dupa: inductia matematica
|
![]()
|
Descriere referat: |
| Metode de ra?ionament I. Metoda direct?( modus poneus) - pornind de la o propozi?ie A ?i folosind principiul silogismului demonstr?m ca o alta propozi?ie este adev?rat?. II. Metoda indirect? – reducerea la absurd ce se bazeaz? pe echivalen?a (p?q) ? (?q??p) III. Metoda induc?iei Metoda induc?iei A. Egalit??i P(n): 1+2+3+...+n=n(n+1)/2 n?1 I. P(1): 1=1 (A) II. P(n) (A)?P(n+1) (A) P(n+1):1+2+3...+n+(n+1)=(n+1)(n+2)/2 n(n+1)/2+(n+1)=(n+1)(n+2)/2 (n+1)(n+2)/2=(n+1)(n+2)/2 (A) I+II? P(n) (A) ?n?1 Deci: 1+2+3...+n = ?k = n(n+1)/2 , n?1 P(n): 1? +2? +3? +...+n?= n(n+1)(2n+1)/6 n?1 I. P(1): 1=1 (A) II. P(n) (A) ?P(n+1) (A) P(n+1): 12 +22 +32 +...+n2+(n+1)2=(n+1)(n+2)(2n+3)/6 n(n+1)(2n+1)/6+(n+1)? =( n+1)(n+2)(2n+3)/6 (n+1)(n+2)(2n+3)/6=( n+1)(n+2)(2n+3)/6 (A) I+II?P(n) (A) ?n?1 12 +22 +32 + ... +n2 = ?k2 = n(n+1)(2n+1)/6, n?1 13+23+33+...+n3= [n(n+1)/2]2 n?1 |
| Nr. | Nume referat | Hits |
| 1 | HDYtlzImNyujZte | 38 |
| 2 | ychHcjleUk | 56 |
| 3 | icFavkpTyJvwx | 304 |
| 4 | IHTgjXQwpsCipv | 299 |
| 5 | XWUiQfrJBuzFjPqhnBm | 315 |
| 6 | Algebra - formule | 1249 |
| 7 | Andrei Dobrescu | 569 |
| 8 | Aplicatii-asemanare | 1229 |
| 9 | BLACK HOLES | 478 |
| 10 | Cercul | 774 |
| 11 | Chestiuni de matematica distractiva | 885 |
| 12 | Concursul interjudetean Pitagora | 613 |
| 13 | Cum rezolvam probleme | 666 |
| 14 | DETERMINANTI TRIGONOMETRICI | 598 |
| 15 | DETERMINAREA CURBELOR | 449 |
| 16 | DETERMINAREA PLANULUI | 702 |
| 17 | DIVIZIBILITATE | 761 |
| 18 | Derivate de ordinul n | 441 |
| 19 | Divizibilitatea numerelor naturale | 595 |
| 20 | Dosar Matematca | 513 |
| 21 | Ecuatii | 526 |
| 22 | Euclid din Alexandria | 415 |
| 23 | Euclid | 507 |
| 24 | FUNCTIA | 466 |
| 25 | Formule de geometrie la matematica(sin si cos) | 583 |