题目内容
已知{an}是正项等比数列,且a1a100=64,求log2a1+log2a2+…+log2a100的值
答案:300
解析:
解析:
| ∵ {an}是正项等比数列
∴ a1a100=a2a99=a3a98=…=a50a51=64 ∴ log2a1+log2a2+…+log2a100 =log2(a1a2…a100)=log2(a1a100)50 =50log2a1a100=50log264=300. |
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