等比数列{an}的首项为1，公比为q，前n项之和为Sn，则1/a1+1/a2+1/a3+....+1/an等于（求过程，要详细！！） 等比数列{an}的首项为1,公比为q,前n项之和为Sn,则数...
\u7b49\u6bd4\u6570\u5217\uff5ban\uff5d\u7684\u9996\u9879\u4e3a1\uff0c\u516c\u6bd4\u4e3aq\uff08q\u4e0d\u7b49\u4e8e1\uff09\uff0c\u524dn\u9879\u548c\u4e3aSn\uff0c\u52191/a1+1/a2+a/a3+...1/an\u7b49\u4e8e\u7b49\u6bd4\u6570\u5217\u7684\u901a\u9879\u516c\u5f0f\u662f\uff1aAn=A1*q^\uff08n\uff0d1\uff09 \u6240\u4ee5 \u9898\u76ee\u4e2d An=q\u7684\uff08n-1\uff09\u6b21\u65b91/a1+1/a2+1/a3+...1/an=1+1/q+1/q\u7684\u5e73\u65b9+\u2026\u2026+1/q\u7684\uff08n-1\uff09\u6b21\u65b9\u5219\u8bbe1/a1\u4e3a\u9996\u9879 bn=\uff081/q\uff09\u7684\uff08n-1\uff09\u6b21\u65b9S\uff08bn\uff09=1/Sn\u5927\u6982\u662f\u8fd9\u4e2a
\u5e0c\u671b\u91c7\u7eb3
\u6574\u7406\u540e\u7684\u7ed3\u679c\u662f\u7ecf\u8fc7\u53d8\u5f62\u7684\uff0cSn'=[a1(1-q^n)/(1-q)]/[(a1^2)q^(n-1)]\uff0c\u52a0\u6df1\u7684\u8fd9\u4e24\u4e2a\u53ef\u4ee5\u6d88\u6389\uff0c\u7136\u540e\u628aSn'=(1/a1)[1-(1/q)^n]/(1-1/q)\u6574\u7406\u4e0b\uff0c\u5c31\u4f1a\u53d1\u73b0\u548c\u6d88\u6389\u4e4b\u540e\u7684\u7ed3\u679c\u4e00\u6837\u4e86
首项为1，公比为q则an=q^(n-1)
所以1/a1+1/a2+1/a3+....+1/an
=1+1/q+1/q^2+...+1/q^(n-1)
=(1-1/q^n)/(1-1/q)
=[1/q^(n-1)]*(q^n-1)/(q-1)
=Sn/q^(n-1)
希望能帮到你O(∩_∩)O
这样的话公比就是1/q 了 带入等比数列求和公式就ok了
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