已知数列{an}的首相a1=2/3,a（n+1）=2an/（an+1）,n=1,2……
2个回答

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

相关问题