| qn−1 | ||
∑n−1k=1 (qk−1) = q | −n | |
| q−1 |
| 1 | ||
Niech a = 1+ | ||
| n |
| 1 | ||
Sn = ∑k=1..n k(1+ | )k = ∑k=1..n kak = [ ∑ xax δx ]x=1x=n+1 | |
| n |
| ax | ax+1 | ax | ax+1 | |||||
∑ xaxδx = x | − ∑ 1* | δx = x | − | + c | ||||
| a−1 | a−1 | a−1 | (a−1)2 |
| ax | ||
= | ((a−1)x − a) + c. | |
| (a−1)2 |
| ax | ||
Sn = [ | ((a−1)x − a) ]x=1x=n+1 | |
| (a−1)2 |
| an+1 | a | |||
= | ((a−1)(n+1) − a) + | |||
| (a−1)2 | (a−1)2 |
| an+1 | a | a | ||||
= | (an − (n+1)) + | = | ||||
| (a−1)2 | (a−1)2 | (a−1)2 |
| x(x−1)(x+1) | p(p+1)(p+2) | |||
∑n=1..p Sn = [ | ]x=1x=p+1 = | |||
| 3 | 3 |
Wszystko jest tam wyjaśnione od początku.