求证:1+1⼀3^2+1⼀5^2+...+1⼀(2n-1)^2>7⼀6-1⼀2(2n+1)

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2024-12-16 14:08:54
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回答1:

证明:
1 + 1 / 3^2 + 1 / 5^2 + … + 1 / (2n-1)^2
= 1 / 2 * [ 1 + 1 / 3^2 + 1 / 5^2 + … + 1 / (2n-1)^2
+ 1 + 1 / 3^2 + 1 / 5^2 + … + 1 / (2n-1)^2 ]
> 1 / 2 * [ 1 + 1 / 3^2 + 1 / 5^2 + … + 1 / (2n-1)^2
+ 1 + 1 / 4^2 + 1 / 6^2 + … + 1 / (2n)^2 ]
= 1 / 2 * [ 2 + 1 / 3^2 + 1 / 4^2 + 1 / 5^2 + … + 1 / (2n)^2 ]
> 1 / 2 * { 2 + 1 / (3 * 4) + 1 / (4 * 5) + … + 1 / [(2n)(2n+1)] }
= 1 / 2 * [ 2 + (1/3 - 1/4) + (1/4 - 1/5) + … + 1/(2n) - 1/(2n+1) ]
= 1 / 2 * [ 2 + 1 / 3 - 1 / (2n+1) ]
= [ 7 / 3 - 1 / (2 n + 1) ] / 2
= 7 / 6 - 1 / [2 (2n+1)]