在各项均为正数的等比数列{an}中，已知a1=1，a2+a3=6，则数列{an}... 在各项均为正数的等比数列{an}中，已知...
\u5728\u5404\u9879\u5747\u4e3a\u6b63\u6570\u7684\u7b49\u6bd4\u6570\u5217{an}\u4e2d\uff0c\u5df2\u77e5a1=1\uff0ca2+a3=6\uff0c\u5219\u6570\u5217{an}\u7684\u901a\u9879\u516c\u5f0f\u4e3a______\u8bbe\u7b49\u6bd4\u6570\u5217\u7684\u516c\u6bd4\u4e3aq\uff0e\u5219\u7531a1=1\uff0ca2+a3=6\uff0c\u5f97\uff1aa1\uff08q+q2\uff09=6?q2+q-6=0\u89e3\u5f97q=2\u6216q=-3\uff0e\u53c8\u56e0\u4e3a\u6570\u5217\u5404\u9879\u5747\u4e3a\u6b63\u6570\u2234q=2\uff0e\u2234an=a1?qn-1=2n-1\uff0e\u6545\u7b54\u6848\u4e3a\uff1aan=2n-1\uff0e
a
\u89e3\uff1a\u8bbe\u7b49\u6bd4\u6570\u5217\u7684\u516c\u6bd4\u4e3aq\uff0e
\u5219\u7531a1=1\uff0ca2+a3=6\uff0c\u5f97\uff1aa1\uff08q+q2\uff09=6⇒q2+q-6=0
\u89e3\u5f97q=2\u6216q=-3\uff0e
\u53c8\u56e0\u4e3a\u6570\u5217\u5404\u9879\u5747\u4e3a\u6b63\u6570
\u2234q=2\uff0e
\u2234an=a1•qn-1=2n-1\uff0e
\u6545\u7b54\u6848\u4e3a\uff1aan=2n-1\uff0e
解答：解：设等比数列的公比为q．
则由a1=1，a2+a3=6，得：a1（q+q2）=6⇒q2+q-6=0
解得q=2或q=-3．
又因为数列各项均为正数
∴q=2．
∴an=a1•qn-1=2n-1．
故答案为：an=2n-1．
点评：本题考查等比数列的基本量之间的关系，若已知等比数列的两项，则等比数列的所有量都可以求出，只要简单数字运算时不出错，问题可解．
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