已知数列{an} an-anan+1-an+1=0 a1=1 求{2n/an}的前n项和 注:an+1的1加在n上

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已知数列{an} an-anan+1-an+1=0 a1=1 求{2n/an}的前n项和 注:an+1的1加在n上
已知数列{an} an-anan+1-an+1=0 a1=1 求{2n/an}的前n项和 注:an+1的1加在n上

已知数列{an} an-anan+1-an+1=0 a1=1 求{2n/an}的前n项和 注:an+1的1加在n上
an-ana(n+1)-a(n+1)=0
an-a(n+1)=ana(n+1)
等式两边同除以ana(n+1)
1/a(n+1)-1/an=1,为定值.
1/a1=1/1=1
数列{1/an}是以1为首项,1为公差的等差数列.
1/an=1+1×(n-1)=n
an=1/n
2n/an=2n/(1/n)=2n²
Sn=2×1/a1+2×2/a2+...+2n/an
=2(1²+2²+...+n²)
=2n(n+1)(2n+1)/6
=n(n+1)(2n+1)/3

an-ana(n+1)-a(n+1)=0
1/a(n+1) -1/an = 1
1/an -1/a1 = n-1
1/an = n
an = 1/n
bn = 2n/an =2
Sn =b1+b2+..+bn = 2n

等式两边同时除以anan+1,得1/an+1 - 1/an=1; 得1/an =n;2n/an =2n^2,,后面就不用写了吧!欢迎追问

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