| 1 | 1 | 1 | 1 | n | |||||
+ | + | +...+ | = | ||||||
| 1*2 | 2*3 | 3*4 | n(n+1) | n+1 |
| n | ||
a jak obliczyc ze ta suma jest rowna | ? | |
| n+1 |
| 1 | 1 | 1 | 1 | |||||
zauważ,że | = | − | = | |||||
| 1*2 | 1 | 2 | 2 |
| 1 | 1 | 1 | 1 | |||||
= | − | = | ||||||
| 2*3 | 2 | 3 | 6 |
| 1 | 1 | 1 | ||||
= | − | |||||
| (n−1)*n | n−1 | n |
| 1 | 1 | 1 | ||||
= | − | |||||
| n(n+1) | n | n+1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | ||||||||
1− | + | − | + ...... + | − | + | − | = po redukcji ... | |||||||
| 2 | 2 | 3 | n−1 | n | n | n+1 |
| 1 | n+1−1 | n | ||||
= 1− | = | = | ||||
| n+1 | n+1 | n+1 |
| 1 | 1 | 1 | |||
= | − | ||||
| n(n+1) | n | n + 1 |
| 1 | 1 | 1 | |||
+ | + | + ... mozna zapisac jako: | |||
| 1*2 | 2*3 | 3*4 |
| 1 | 1 | 1 | 1 | 1 | 1 | |||||||
( | − | ) + ( | − | ) + ( | − | ), co po redukcji (dla n = 3) | ||||||
| 1 | 2 | 2 | 3 | 3 | 4 |
| 1 | 1 | n | ||||
1 − | , czyli ogolnie przyjmie postac: 1 − | = | ||||
| 4 | n + 1 | n + 1 |
| 1 | 1 | 1 | |||
+ | + | <0 | |||
| x(x+1) | (x+1)(x+2) | (x+2)(x+3) |
| 1 | 1 | |||
∑ ( | − | )= | ||
| k | k+1 |
| 1 | 1 | 1 | 1 | 1 | 100 | |||||||
= ∑ | = | + | + | +...+ | = | dobrze? | ||||||
| k(k+1) | 2 | 6 | 12 | 10100 | 101 |