关于等比数列裂项求和法.这是怎么推导出来的?

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/13 01:37:47
关于等比数列裂项求和法.这是怎么推导出来的?

关于等比数列裂项求和法.这是怎么推导出来的?
关于等比数列裂项求和法.

这是怎么推导出来的? 

关于等比数列裂项求和法.这是怎么推导出来的?
an=1/(4n+1)(4n+3)
=1/2[(4n+3)--(4n+1)]/(4n+1)(4n+3)
=1/2[(4n+3)/(4n+1)(4n+3)--(4n+1)/(4n+1)(4n+3)]
=1/2[1/(4n+1)--1/(4n+3)].
an=1/(根号a+根号b)=(根号a--根号b)/(根号a+根号b)(根号a--根号b)
=(根号a--根号b)/(a--b).

1) an=1/(4n+1)(4n+3)=1/2*(4n+3-4n-1)/(4n+1)(4n+3)=1/2*[1/(4n+1)-1/(4n+3)]
2) an=1/(√a+√b)=(√a-√b)/[(√a+√b)(√a-√b)]=(√a-√b)/(a+b)


(1)∵ (4n+3)-(4n+1)=2
∴ 1/[(4n+1)(4n+3)]
=(1/2)*2/[(4n+1)(4n+3)]
=(1/2)*[(4n+3)-(4n+1)]/[(4n+1)(4n+3)]
=(1/2)*[(4n+3)/[(4n+1)(4n+3)]-(4n+1)/[(4n+1)(4n+3)]]
=(1/2)[1/(4n+1)-1/(4n+3)]
(2)分母有理化即可,分子分母同时乘以√a-√b
即得 an=[1/(a-b)]*(√a-√b)