f[0]=1
by Happy_Fool @ 2024-04-19 18:56:14
@[Happy_Fool](/user/1330396)
[Fi,1] = [Fi-1,1]*
2,0
1,1
by Qshuqi @ 2024-04-19 19:05:09
这个好像可以直接推方程吧
$f[n]=2^{n-1}+2^{n-2}+...+2^0$
$=2^{n-1}-1$
by absolute_value @ 2024-04-19 19:05:29
@[Qshuqi](/user/547616) @[absolute_value](/user/1067907)
谢谢！
by Happy_Fool @ 2024-04-19 19:09:33
$$
\begin{bmatrix}
2&1\\
0&1
\end{bmatrix}
\cdot
\begin{bmatrix}
f_i\\
1
\end{bmatrix}
=
\begin{bmatrix}
f_{i+1}\\
1
\end{bmatrix}
$$
核心思路是将 $1$ 扔入转移矩阵。然后根据递推式直接写。
$$
\begin{bmatrix}
2&1\\
0&1
\end{bmatrix}^n
\cdot
\begin{bmatrix}
1\\
1
\end{bmatrix}
=
\begin{bmatrix}
f_{n}\\
1
\end{bmatrix}
$$
by yukimianyan @ 2024-04-19 19:10:22
楼上的楼上的楼上正解
by Max__universe @ 2024-04-19 19:17:36
@[yukimianyan](/user/509229) 谢谢！
by Happy_Fool @ 2024-04-19 19:35:59