已知各项均为正数的等比数列{an}的前n项和为Sn,a1=5,S3=155,求数列{an}的通项公式 已知各项均为正数的等比数列{an}的首项a1=2，Sn为其前...
\u5df2\u77e5\u5404\u9879\u5747\u4e3a\u6b63\u6570\u7684\u7b49\u6bd4\u6570\u5217{an}\uff0c\u9996\u9879a1=12\uff0c\u524dn\u9879\u548c\u4e3aSn\uff0c\u4e14S3+a3\uff0cS5+a5\uff0cS4+a4\u6210\u7b49\u5dee\u6570\u5217\uff0e\uff08\u2160\uff09\u6c42\uff1a\uff08\u2160\uff09\u8bbe\u6b63\u9879\u7b49\u6bd4\u6570\u5217{an}\uff08n\u2208N*\uff09\u7684\u516c\u6bd4\u4e3aq\uff08q\uff1e0\uff09\uff0c\u53c8a1=12\uff0c\u2234an=12?qn-1\uff0c\u2235S3+a3\u3001S5+a5\u3001S4+a4\u6210\u7b49\u5dee\u6570\u5217\uff0c\u22342\uff08S5+a5\uff09=\uff08S3+a3\uff09+\uff08S4+a4\uff09\uff0c\u53732\uff08a1+a2+a3+a4+2a5\uff09=\uff08a1+a2+2a3\uff09+\uff08a1+a2+a3+2a4\uff09\uff0c\u5316\u7b80\u5f974a5=a3\uff0c\u22344a1q4=a1q2\uff0c\u5316\u4e3a4q2=1\uff0c\u89e3\u5f97q=\u00b112\u2235q\uff1e0\uff0c\u2234q=12\uff0c\u2234an=12n\uff08 II\uff09\u7531\uff08 I\uff09\u77e5\uff0cnan=n2n\uff0c\u5219Tn=12+222+323+\u2026+n2n\uff0c\u246012Tn=122+223+324+\u2026+n?12n+n2n+1\uff0c\u2461\u2026\uff088\u5206\uff09\u2460-\u2461\u5f97\uff1a12Tn=12+122+\u2026+12n-n2n+1=12(1?12n)1?12-<td style="border-bo
\uff081\uff09\u22355S1\uff0cS3\uff0c3S2\u6210\u7b49\u5dee\u6570\u5217\uff0c\u22342S3=5S1+3S2\u2026\uff081\u5206\uff09\u53732\uff08a1+a1q+a1q2\uff09=5a1+3\uff08a1+a1q\uff09\uff0c\u5316\u7b80\u5f97 2q2-q-6=0\u2026\uff082\u5206\uff09\u89e3\u5f97\uff1aq=2\u6216q=-32\u2026\uff083\u5206\uff09\u56e0\u4e3a\u6570\u5217{an}\u7684\u5404\u9879\u5747\u4e3a\u6b63\u6570\uff0c\u6240\u4ee5q=-32\u4e0d\u5408\u9898\u610f\u2026\uff084\u5206\uff09\u6240\u4ee5{an}\u7684\u901a\u9879\u516c\u5f0f\u4e3a\uff1aan=2n\uff0e\u2026\uff085\u5206\uff09\uff082\uff09\u7531bn=log2an\u5f97bn=log22n=n\u2026\uff086\u5206\uff09\u2234cn=1bnbn+1=1n(n+1)=1n-1n+1\u2026\uff087\u5206\uff09\u2234Tn=1-12+12-13+\u2026+1n-1n+1=1?1n+1=nn+1\u2026\uff088\u5206\uff09\u2235nn+1\u2264k\uff08n+4\uff09\u2234k\u2265n(n+1)(n+4)=nn2+5n+4\u2026\uff089\u5206\uff09=1n+4n+5\u2026-\uff0811\u5206\uff09\u2235n+4n+5\u22652n?4n+5=9\uff0c\u5f53\u4e14\u4ec5\u5f53n=4n\uff0c\u5373n=2\u65f6\u7b49\u53f7\u6210\u7acb------\uff0812\u5206\uff09\u22341n+<table cellpadding="-1" cellspacing="-1" style="
解：等比数列各项均为正数，公比q>0
S3=a1+a1q+a1q²=a1(1+q+q²)=5(1+q+q²)=155
q²+q+1=31
q²+q-30=0
(q+6)(q-5)=0
q=-6(<0，舍去)或q=5
an=a1q^(n-1)=5ⁿ
n=1时，a1=5，同样满足。
数列{an}的通项公式为an=5ⁿ。
等比数列{an} 又a1=5 那么
S3＝155＝5（1－q`3）/（1－q） 化简得q`3－31q＋30＝（q－1）（q`2＋q－30）＝0
推出 q1＝1 q2＝5 q3＝ -6（舍，题目提示的：各项均为正数的等比数列）
当q＝1 ，S3＝3×a1＝15 与题目当中的 S3＝155 不符合。所以只限于 n＝1 的情况
所以数列{an}的通项公式为
an=5 × 5`（n－1） ＝ 5`n （n≥2）
a1＝5 （n＝1）
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