【已知等差数列{an}的前n项和为Sn,且(2n-1)Sn+1-(2n+1)Sn=4namp;#178;-1(n∈N*)求证：1/√a1+1/√a2+.+1/√an＞1/2(√4n+1-1)】
<p>问题：【已知等差数列{an}的前n项和为Sn,且(2n-1)Sn+1-(2n+1)Sn=4namp;#178;-1(n∈N*)求证：1/√a1+1/√a2+.+1/√an＞1/2(√4n+1-1)】<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">李京峰的回答：<div class="content-b">网友采纳 Sn+1/(2n+1)-Sn/(2n-1)=1Sn/(2n-1)=S1+n-1→Sn=(S1+n-1)(2n-1)→Sn=n(2n-1)an=4n-31/√an=2/2√(4n-3)>2/(√4n-3+√4n+1)=(√4n+1-√4n-3)/21/√a1+1/√a2+.+1/√an>(√5-√1+√9-√5+√13-√9+...√4n+1-√4n-3...
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