| 2n + 3n | ||
Zbadać monotoniczność ciągu an= | . | |
| 2n+1 + 3n+1 |
| 2n+3n | ||
lekka pomyłka ma być an= | ||
| 2n+1 + 3n+1 |
| 2*2n + 3*3n | ||
an+1 = | ||
| 2*2n+1 + 3*3n+1 |
| (2n*3n*3)(2n*2+3n*3) | ||
doszedłem do | − | |
| (2n2*2n+3n2*3n)(2n*2+3n*3) |
| (2n+3n)(2n2*2n+3n2*3n) | ||
i nie wiem co z tymi | ||
| (2n2*22+3n2*3n)(2n*2+3n*3) |
| 1 | 1 | |||
an= | (1 − | ) | ||
| 2 | 2(2/3)n+3 |
| an+1 | |
niż an+1−an | |
| an |
| 1 + un | ||
an = 1/2 | gdzie u = 3/2 | |
| 1 + un + 1 |
| 1 + un+1 | 1 + un | |||
2[an + 1 − an] = | − | = | ||
| 1 + un + 2 | 1 + un + 1 |
| (1 + un+1)2 − (1 + un)(1 + un + 2) | ||
= | = | |
| (1 + un + 1)(1 + un + 2) |
| 2un + 1 − un − un + 2 | −u(un + 1 − un) + 1(un + 1 − un) | |||
= | = | = | ||
| ... | ... |
| (1 − u)(un + 1 − un) | ||
= | < 0 | |
| ... |