n+1 |
n+2 |
设{an}的前n项和为Sn=log2
2 |
3 |
3 |
4 |
n |
n+1 |
n+1 |
n+2 |
=[log22-log23]+[log23-log24]+…+[log2n-log2(n+1)]+[log2(n+1)-log2(n+2)]
=[log22-log2(n+2)]=log2
2 |
n+2 |
即
2 |
n+2 |
解得n+2>64,
n>62;
∴使Sn<-5成立的自然数n有最小值为63.
故选:A.
n+1 |
n+2 |
n+1 |
n+2 |
2 |
3 |
3 |
4 |
n |
n+1 |
n+1 |
n+2 |
2 |
n+2 |
2 |
n+2 |