数列满足 b(1)=0，b(2)=-ln2，2b(n+2)-b(n+1)-b(n)=0，n=1,2,…。求 b(n) 的通项公式

wintex

請問數列

王守恩

加深影响。

\(\displaystyle b(n)=0,\ln(2),\frac{\ln(2)}{2},\frac{3\ln(2)}{4},\frac{5\ln(2)}{8},\frac{11\ln(2)}{16},\frac{21\ln(2)}{32},\frac{43\ln(2)}{64},\frac{85\ln(2)}{128},\frac{171\ln(2)}{256},\frac{341\ln(2)}{512},\)

LinearRecurrence[{1, 2}, {0, 1}, 28]    或\(\displaystyle\bigg[\frac{2^n}{3}\bigg]\)
{0, 1, 1, 3, 5, 11, 21, 43, 85, 171, 341, 683, 1365, 2731, 5461, 10923, 21845, 43691, 87381,
174763, 349525, 699051, 1398101, 2796203, 5592405, 11184811, 22369621, 44739243,}

\(\displaystyle\lim_{n\to\infty}\frac{\frac{2^{n+1}+\cos(n\pi)}{3}\ln(2)}{2^n}=\frac{2\ln(2)}{3}=0.46209812\)

luyuanhong