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While thinking this problem,I found an identity: $b_n^TH_n^{-1}b_n+\frac{1}{n^2}=2$,where $b_n=(\frac{1}{1^2} \frac{1}{2^2} ... \frac{1}{n^2})^T$,and $H_n$ is the n-th Hilbert matrix.However,I cannot prove it directly.Does anybody give me some hints to prove this identity?

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