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

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
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

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
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)

an=(2n)^2/(2n+1)(2n-1)
=2n/(2n-1) *2n/(2n+1)
=(1+1/2n-1)*(1-1/2n+1)
=1+1/2n-1 -1/2n+1- 1/(2n-1)(2n+1)
=1+1/2n-1 -1/2n+1-1/2*[1/2n-1 -1/2n+1]
=1+1/2*[1/2n-1 -1/2n+1]
Sn=n+1/2*[...

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an=(2n)^2/(2n+1)(2n-1)
=2n/(2n-1) *2n/(2n+1)
=(1+1/2n-1)*(1-1/2n+1)
=1+1/2n-1 -1/2n+1- 1/(2n-1)(2n+1)
=1+1/2n-1 -1/2n+1-1/2*[1/2n-1 -1/2n+1]
=1+1/2*[1/2n-1 -1/2n+1]
Sn=n+1/2*[1/1-1/3+1/3-1/5+..+1/2n-1 -1/2n+1]
=n+1/2*[1-1/(2n+1)]
=n+1/2-1/2*(2n+1)
=[4n^2+2n+2n+1-1]/[2*(2n+1)]
=2[n^2+n]/(2n+1)
=2n*(n+1)/(2n+1)

收起

Sn=2^/1*3+4^2/3*5+……(2n)^2/(2n-1)(2n+1)
=1+1/1*3+1+1/3*5+...+1+1/(2n-1)(2n+1)
=n+(1/1*3+1/3*5+...+1/(2n-1)(2n+1))
=n+[(1-1/3)/2+(1/3-1/5)/2+...+(1/(2n-1)-1/(2n+1)/2]
=n+(1-1/(2n+1))/2
=n+(2n/(2n+1)/2
=n+n/(2n+1)
=(2n^2+2n)/(2n+1)
=2n(n+1)/(2n+1)

An=(2n)^2/(2n+1)(2n-1)
=[(2n+1)(2n-1)+1]/(2n+1)(2n-1)
=1+1/(2n+1)(2n-1)
=1+[1/(2n-1)-1/(2n+1)]/2
所以Sn=1+（1/1-1/3）/2 +1+[1/3-1/5]/2 +1+[1/(2n-1)-1/(2n+1)]/2
=n+[1/1-1/3+1/3+1/3+....+1/(2n-1)-1/(2n+1)]/2
=n+[1-1/(2n+1)]/2
=2n(n+1)/(2n+1)

An=(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)=1-1/2[1/(2n-1)-1/(2n+1)]
SN=N-1/2∑[1/(2n-1)-1/(2n+1)]
答案出来了

Sn=1/2{2^2(1/1-1/3)+4^2(1/3-1/5)+.......+(2n)^2[1/(2n-1)-1/(2n+1)]}
2Sn=[2^2+4^2/3+.....+(2n)^2/(2n-1)]-[2^2/3+4^2/5+......+(2n)^2/(2n+1)]
设An=2^2+4^2/3+.....+(2n)^2/(2n-1)
Bn=2^2/3+4^2/5+......+(2n)^2/(2n+1)

因为：1/(2n+1)(2n-1) = 1/2 * [1/(2n - 1) - 1/(2n + 1)]
所以：
(2n)^2/(2n+1)(2n-1)
= (4n^2 / 2) * [1/(2n - 1) - 1/(2n + 1)]
= 2n^2 / (2n - 1) - 2n^2 / (2n + 1)
又因为：
2(n+1)^2 / [2*(n+1...

全部展开

因为：1/(2n+1)(2n-1) = 1/2 * [1/(2n - 1) - 1/(2n + 1)]
所以：
(2n)^2/(2n+1)(2n-1)
= (4n^2 / 2) * [1/(2n - 1) - 1/(2n + 1)]
= 2n^2 / (2n - 1) - 2n^2 / (2n + 1)
又因为：
2(n+1)^2 / [2*(n+1) - 1] - 2n^2 / (2n + 1)
= (2n^2 + 4n + 2) / (2n + 1) - 2n^2 / (2n + 1)
= (4n + 2)/(2n + 1)
= 2
所以：
Sn = 2*(1^2)/1 - 2*(1^2)/3 + 2*(2^2)/3 - 2*(2^2)/5 +...+2*(n^2)/(2n-1) - 2*(n^2)/(2n+1)
= 2*(1^2)/1 + [2*(2^2)/3 - 2*(1^2)/3] + ... + [2*(n^2)/(2n-1) - 2*(n-1)^2/[2*(n-1) +1]] - 2*(n^2)/(2n+1)
= 2 *(1^2)/1 + 2 + 2 + 2 + .... - 2*(n^2)/(2n+1) [中间有n-1个2]
= 2n - 2n^2/(2n+1)
= (2n^2 + 2n)/(2n + 1)
= 2n(n + 1)/(2n + 1)

收起