数学计算题？
3个回答
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an
= 4n^2/[(2n-1)(2n+1)]
=1 + 1/[(2n-1)(2n+1)]
=1 + (1/2)[ 1/(2n-1) -1/(2n+1) ]
Sn
=a1+a2+...+an
=n + (1/2)[ 1 -1/(2n+1) ]
=n+ n/(2n+1)
(2x2)/(1x3)+ (4x4)/(3x5)+...+(2024x2024)/(2023x2025)
=S1012
=1012 + 1012/(2025)
=1011 又 2025分之1012
= 4n^2/[(2n-1)(2n+1)]
=1 + 1/[(2n-1)(2n+1)]
=1 + (1/2)[ 1/(2n-1) -1/(2n+1) ]
Sn
=a1+a2+...+an
=n + (1/2)[ 1 -1/(2n+1) ]
=n+ n/(2n+1)
(2x2)/(1x3)+ (4x4)/(3x5)+...+(2024x2024)/(2023x2025)
=S1012
=1012 + 1012/(2025)
=1011 又 2025分之1012
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展开全部
an
= 4n^2/[(2n-1)(2n+1)]
=1 + 1/[(2n-1)(2n+1)]
=1 + (1/2)[ 1/(2n-1) -1/(2n+1) ]
Sn
=a1+a2+...+an
=n + (1/2)[ 1 -1/(2n+1) ]
=n+ n/(2n+1)
(2x2)/(1x3)+ (4x4)/(3x5)+...+(2024x2024)/(2023x2025)
=S1012
=1012 + 1012/(2025)
=1012 又 2025分之1012
= 4n^2/[(2n-1)(2n+1)]
=1 + 1/[(2n-1)(2n+1)]
=1 + (1/2)[ 1/(2n-1) -1/(2n+1) ]
Sn
=a1+a2+...+an
=n + (1/2)[ 1 -1/(2n+1) ]
=n+ n/(2n+1)
(2x2)/(1x3)+ (4x4)/(3x5)+...+(2024x2024)/(2023x2025)
=S1012
=1012 + 1012/(2025)
=1012 又 2025分之1012
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