作业帮求和Sn=2^2/1*3+4^2/3*5+………+(2n)^2/(2n+1)(2n-1)
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求和Sn=2^2/1*3+4^2/3*5+………+(2n)^2/(2n+1)(2n-1)
结果Sn=2n(n+1)/(2n+1)是这个结果的请把答案发上来
结果Sn=2n(n+1)/(2n+1)是这个结果的请把答案发上来
简化通项
(2n)^2/(2n+1)(2n-1)=4n^2/(4n^2-1)=(4n^2-1+1)/(4n^2-1)=1+1/(4n^2-1)
=1+1/(2n+1)(2n-1)
且
1/(2n+1)(2n-1)=[1/(2n-1)-1/(2n+1)]/2
所以原式子=1+1+1+1...+1+1/2[1-1/3+1/3-1/5+1/5-.+1/2n-1-1/2n+1]
=n+1/2(1-1/2n+1)
=n+n/(2n+1)
=2n(n+1)/(2n+1)
(2n)^2/(2n+1)(2n-1)=4n^2/(4n^2-1)=(4n^2-1+1)/(4n^2-1)=1+1/(4n^2-1)
=1+1/(2n+1)(2n-1)
且
1/(2n+1)(2n-1)=[1/(2n-1)-1/(2n+1)]/2
所以原式子=1+1+1+1...+1+1/2[1-1/3+1/3-1/5+1/5-.+1/2n-1-1/2n+1]
=n+1/2(1-1/2n+1)
=n+n/(2n+1)
=2n(n+1)/(2n+1)
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