已知数列{an}的首相a1=2/3,a(n+1)=2an/(an+1),n=1,2……
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已知数列{an}的首相a1=2/3,a(n+1)=2an/(an+1),n=1,2……
(1)证明:数列{(1/an)-1}是等比数列.
(2)求数列{n/an}的前n项和Sn.
(1)证明:数列{(1/an)-1}是等比数列.
(2)求数列{n/an}的前n项和Sn.
![已知数列{an}的首相a1=2/3,a(n+1)=2an/(an+1),n=1,2……](/uploads/image/z/16795126-46-6.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%A6%96%E7%9B%B8a1%3D2%2F3%2Ca%EF%BC%88n%2B1%EF%BC%89%3D2an%2F%EF%BC%88an%2B1%EF%BC%89%2Cn%3D1%2C2%E2%80%A6%E2%80%A6)
1)证明:a(n+1)=2an/(an + 1)
1/a(n+1)=1/2+(1/2)*(1/an)
1/a(n+1)-1=(1/2)*(1/an-1)
1/an -1=a1*(1/2)^(n-1)=(3/2-1)*(1/2)^(n-1)=(1/2)*(1/2)^(n-1)
即数列{1/an - 1}是q=1/2的等比数列
令bn=n/an=n[1+(2/3)*(1/2)^(n-1)]
Sn=(i=1~n)Σbi= (i=1~n)Σi + (i=1~n)Σ(1/2)*i* (1/2)^(i-1)
令f(x)= (i=1~n)Σ x^i = x*(x^n-1)/(x-1)=[x^(n+1)-x]/(x-1)
f'(x)= (i=1~n)Σ i*x^(i-1) = [nx^(n+1)-(n+1)x^n+1]/(x-1)^2
Sn=n(n+1)/2 + (1/2)*[n(1/2)^(n+1) - (n+1)(1/2)^n + 1]/(1/2-1)^2
=(n^2+n+4)/2-(n+2)/2^n
1/a(n+1)=1/2+(1/2)*(1/an)
1/a(n+1)-1=(1/2)*(1/an-1)
1/an -1=a1*(1/2)^(n-1)=(3/2-1)*(1/2)^(n-1)=(1/2)*(1/2)^(n-1)
即数列{1/an - 1}是q=1/2的等比数列
令bn=n/an=n[1+(2/3)*(1/2)^(n-1)]
Sn=(i=1~n)Σbi= (i=1~n)Σi + (i=1~n)Σ(1/2)*i* (1/2)^(i-1)
令f(x)= (i=1~n)Σ x^i = x*(x^n-1)/(x-1)=[x^(n+1)-x]/(x-1)
f'(x)= (i=1~n)Σ i*x^(i-1) = [nx^(n+1)-(n+1)x^n+1]/(x-1)^2
Sn=n(n+1)/2 + (1/2)*[n(1/2)^(n+1) - (n+1)(1/2)^n + 1]/(1/2-1)^2
=(n^2+n+4)/2-(n+2)/2^n
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