,故点P在B1A1的延长线上,且A1P
4.解:⑴当n 1时,a1 3.
假设当n k时,ak 4mk 3,mk N
121. 2
a4m 34m 3
则当n k 1时,ak 1 3k 3k (4 1)k
4mk 30
C4 ( 1)0 mk 344mk 30C4)4mk 3 mk 34 ( 1
4mk 21
C4 ( 1)1 mk 34
…
4mk 21
C4)4mk 2 mk 34 ( 1
4T 1 4(T 1) 3
4m 24m 14m 24m 20011*
其中T C4mk 34k ( 1) C4mk 34k ( 1) … C4mkk 3 ( 1)k N.
所以 mk 1 T 1 N,使 ak 1 4mk 1 3所以当n k+1时,结论也成立, 所以 n N*, mn N,使an 4mn 3; (2)an 1 3
an
34mn 3 (81)mn 27,故a2010的末位数字是7.