Question

已知数列{an}满足:a1=3/4,且对任意正整数n,有1/an+1=1/2(1/an十1)

Answer 1

解:
1/a(n+1)=1/2(1/an +1)
2(an + 1)=a(n+1)
2an + 4=a(n+1)+2
2(an+ 2)=a(n+1) +2
2=[a(n+1) +2]/(an+ 2)
所以
{an + 2}为等比数列
首项a1 +2=3/4+ 2=11/4
公比q=2
所以
an +2=11/4(2)^(n-1)
所以
an=11(2)^(n-3) - 2

Answer 2

1/a(n+1)=(1/2)(1/an+1)
1/a(n+1) -1=(1/2)(1/an -1)
{1/an -1} 是等比数列, q=1/2
1/an-1 = (1/2)^(n-1) .(1/a1-1)
=(1/3)(1/2)^(n-1)
1/an = 1+(1/3)(1/2)^(n-1)
an = 1/1+(1/3)(1/2)^(n-1)]