数列 (an+1)=(an^2)/(1+2an) 怎么求通项公式?
2个回答
a(n+1)=[a(n)]^2/[1+2a(n)],
若a(1)=0,则,a(n)=0,n=1,2,...
若a(1)不等于0,则a(n)不等于0,
1/a(n+1)=[1+2a(n)]/[a(n)]^2=1/[a(n)]^2 + 2/a(n),
1/a(n+1) + 1 = [1/a(n) + 1]^2,
若a(1)=-1,则a(n)=-1,n=1,2,...
若a(1)不等于-1,a(1)不等于0,则,
n>1时, 1/a(n) + 1 > 0,
ln[1/a(n+1) + 1] = 2ln[1/a(n) + 1],
{ln[1/a(n) + 1]}是首项为ln|1/a(1) + 1|, 公比为2的等比数列.
ln[1/a(n) + 1] = ln|1/a(1) + 1| 2^(n-1),
1/a(n) + 1 = |1/a(1) + 1|e^[2^(n-1)],
1/a(n) = |1/a(1) + 1|e^[2^(n-1)] - 1,
a(n)={|1/a(1)+1|e^[2^(n-1)] - 1}^(-1),n=1,2,...
相关问题
