
设数列{an}满足a1=2,an+1=an+1/an(n=1,2,3......),证明:an>根号下(2n+1).急用
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证明:(开方即 1/2 次方,用^表示)
n=1时,A_1 = 4^(1/2) = 2 > (2*1 + 1)^(1/2) = 3^(1/2);
假设n=k时,A_k > (2k + 1)^(1/2);
则n=(k+1)时,A_(k+1) = A_k + 1/A_k;
欲证结果,只需证(A_k + 1/A_k)^2 > (2k + 3);而
(A_k + 1/A_k)^2 = (A_k)^2 + 2 + 1/[(A_k)^2] > 2k + 1 + 2 + 1/[(A_k)^2] = (2k + 3) + 1/[(A_k)^2];
易知A_k > 0且单调递增。因而1/[(A_k)^2] > 0;
所以,(A_k + 1/A_k) > (2k + 3)^(1/2)
n=1时,A_1 = 4^(1/2) = 2 > (2*1 + 1)^(1/2) = 3^(1/2);
假设n=k时,A_k > (2k + 1)^(1/2);
则n=(k+1)时,A_(k+1) = A_k + 1/A_k;
欲证结果,只需证(A_k + 1/A_k)^2 > (2k + 3);而
(A_k + 1/A_k)^2 = (A_k)^2 + 2 + 1/[(A_k)^2] > 2k + 1 + 2 + 1/[(A_k)^2] = (2k + 3) + 1/[(A_k)^2];
易知A_k > 0且单调递增。因而1/[(A_k)^2] > 0;
所以,(A_k + 1/A_k) > (2k + 3)^(1/2)
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