3 | ||
zauważmy dodatkowo, że 3ak = 2bk ... czyli bk = | ak | |
2 |
3 | ||
= 10*c2k + 90bk2 + 60* | ak + 11 = | |
2 |
2 | 2 | |||
22= | (102−1) to 22......2(n dwójek) = | (10n−1) | ||
9 | 9 |
3 | 1 | |||
n trójek 33.....3 = | (10n−1) to (33....3)2= | (10n−1)2 | ||
9 | 9 |
1 | ||
2n jedynek 11.....1= | (102n−1) | |
9 |
1 | 1 | |||
L= | (2*10n−2+102n−2*10n+1)= | (102n−1)= 11.... 1 −−− 2n jedynek | ||
9 | 9 |