已知等差数列{an}的各项均为正数 求证:1/(√a1+√a2)+1/(√a2+√a3)+……+1/(√an-1+√an)=(已知等差数列{an}的各项均为正数 求证:1/(√a1+√a2)+1/(√a2+√a3)+……+1/(√an-1+√an)=(n-1)/(√a1+√an)
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![已知等差数列{an}的各项均为正数 求证:1/(√a1+√a2)+1/(√a2+√a3)+……+1/(√an-1+√an)=(已知等差数列{an}的各项均为正数 求证:1/(√a1+√a2)+1/(√a2+√a3)+……+1/(√an-1+√an)=(n-1)/(√a1+√an)](/uploads/image/z/5528681-17-1.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%90%84%E9%A1%B9%E5%9D%87%E4%B8%BA%E6%AD%A3%E6%95%B0+%E6%B1%82%E8%AF%81%EF%BC%9A1%2F%28%E2%88%9Aa1%2B%E2%88%9Aa2%29%2B1%2F%28%E2%88%9Aa2%2B%E2%88%9Aa3%29%2B%E2%80%A6%E2%80%A6%2B1%2F%28%E2%88%9Aan-1%2B%E2%88%9Aan%29%3D%EF%BC%88%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%90%84%E9%A1%B9%E5%9D%87%E4%B8%BA%E6%AD%A3%E6%95%B0+%E6%B1%82%E8%AF%81%EF%BC%9A1%2F%28%E2%88%9Aa1%2B%E2%88%9Aa2%29%2B1%2F%28%E2%88%9Aa2%2B%E2%88%9Aa3%29%2B%E2%80%A6%E2%80%A6%2B1%2F%28%E2%88%9Aan-1%2B%E2%88%9Aan%29%3D%EF%BC%88n-1%EF%BC%89%2F%28%E2%88%9Aa1%2B%E2%88%9Aan%29)
已知等差数列{an}的各项均为正数 求证:1/(√a1+√a2)+1/(√a2+√a3)+……+1/(√an-1+√an)=(已知等差数列{an}的各项均为正数 求证:1/(√a1+√a2)+1/(√a2+√a3)+……+1/(√an-1+√an)=(n-1)/(√a1+√an)
已知等差数列{an}的各项均为正数 求证:1/(√a1+√a2)+1/(√a2+√a3)+……+1/(√an-1+√an)=(
已知等差数列{an}的各项均为正数 求证:1/(√a1+√a2)+1/(√a2+√a3)+……+1/(√an-1+√an)=(n-1)/(√a1+√an)
已知等差数列{an}的各项均为正数 求证:1/(√a1+√a2)+1/(√a2+√a3)+……+1/(√an-1+√an)=(已知等差数列{an}的各项均为正数 求证:1/(√a1+√a2)+1/(√a2+√a3)+……+1/(√an-1+√an)=(n-1)/(√a1+√an)
1/(√an-1+√an)=(√an-1-√an)/(√an-1+√an)(√an-1-√an)
=-1/d(√an-1-√an) d为等差数列{an}的的公差
左边=-1/d(√a1-√a2+√a2-√a3+……+√an-1-√an)
=-1/d(√a1-√an)
=(√an-√a1)/d
= (an-a1)/d(√a1+√an)
=(n-1)/(√a1+√an)=右边
所以 原等式成立