斐波那契数
為求得斐波那契數列的一般表達式，可以藉助線性代數的方法。高中的初等數學知識也能求出。
初等代數解法
编辑
已知：
a
1
=
1
{\displaystyle a_{1}=1}
a
2
=
1
{\displaystyle a_{2}=1}
a
n
=
a
n
−
1
+
a
n
−
2
{\displaystyle a_{n}=a_{n-1}+a_{n-2}}
（n≥3）
首先構建等比數列
编辑
設
a
n
+
α
a
n
−
1
=
β
(
a
n
−
1
+
α
a
n
−
2
)
{\displaystyle a_{n}+\alpha a_{n-1}=\beta (a_{n-1}+\alpha a_{n-2})}
化簡得
a
n
=
(
β
−
α
)
a
n
−
1
+
α
β
a
n
−
2
{\displaystyle a_{n}=(\beta -\alpha )a_{n-1}+\alpha \beta a_{n-2}}
比較係數可得：
{
β
−
α
=
1
α
β
=
1
{\displaystyle {\begin{cases}\beta -\alpha =1\\\alpha \beta =1\end{cases}}}
不妨設
β
>
0
,
α
>
0
{\displaystyle \beta >0,\alpha >0}
解得：
{
α
=
5
−
1
2
β
=
5
+
1
2
{\displaystyle {\begin{cases}\alpha ={\dfrac {{\sqrt {5}}-1}{2}}\\\beta ={\dfrac {{\sqrt {5}}+1}{2}}\end{cases}}}
又因为有
a
n
+
α
a
n
−
1
=
β
(
a
n
−
1
+
α
a
n
−
2
)
{\displaystyle a_{n}+\alpha a_{n-1}=\beta (a_{n-1}+\alpha a_{n-2})}
，
即
{
a
n
+
α
a
n
−
1
}
{\displaystyle \left\{a_{n}+\alpha a_{n-1}\right\}}
為等比數列。
求出數列{an+αan-1}
编辑
由以上可得：
a
n
+
1
+
α
a
n
=
(
a
2
+
α
a
1
)
β
n
−
1
=
(
1
+
α
)
β
n
−
1
=
β
n
{\displaystyle {\begin{aligned}a_{n+1}+\alpha a_{n}&=(a_{2}+\alpha a_{1})\beta ^{n-1}\\&=(1+\alpha )\beta ^{n-1}\\&=\beta ^{n}\\\end{aligned}}}
變形得：
a
n
+
1
β
n
+
1
+
α
β
⋅
a
n
β
n
=
1
β
{\displaystyle {\frac {a_{n+1}}{\beta ^{n+1}}}+{\frac {\alpha }{\beta }}\cdot {\frac {a_{n}}{\beta ^{n}}}={\frac {1}{\beta }}}
。
令
b
n
=
a
n
β
n
{\displaystyle b_{n}={\frac {a_{n}}{\beta ^{n}}}}
求數列{bn}進而得到{an}
编辑
b
n
+
1
+
α
β
b
n
=
1
β
{\displaystyle b_{n+1}+{\frac {\alpha }{\beta }}b_{n}={\frac {1}{\beta }}}
設
b
n
+
1
+
λ
=
−
α
β
(
b
n
+
λ
)
{\displaystyle b_{n+1}+\lambda =-{\frac {\alpha }{\beta }}(b_{n}+\lambda )}
，解得
λ
=
−
1
α
+
β
{\displaystyle \lambda =-{\frac {1}{\alpha +\beta }}}
。
故數列
{
b
n
+
λ
}
{\displaystyle \left\{b_{n}+\lambda \right\}}
為等比數列
即
b
n
+
λ
=
(
−
α
β
)
n
−
1
(
b
1
+
λ
)
{\displaystyle b_{n}+\lambda =\left(-{\frac {\alpha }{\beta }}\right)^{n-1}\left(b_{1}+\lambda \right)}
。而
b
1
=
a
1
β
=
1
β
{\displaystyle b_{1}={\frac {a_{1}}{\beta }}={\frac {1}{\beta }}}
，
故有
b
n
+
λ
=
(
−
α
β
)
n
−
1
(
1
β
+
λ
)
{\displaystyle b_{n}+\lambda =\left(-{\frac {\alpha }{\beta }}\right)^{n-1}\left({\frac {1}{\beta }}+\lambda \right)}
又有
{
α
=
5
−
1
2
β
=
5
+
1
2
{\displaystyle {\begin{cases}\alpha ={\dfrac {{\sqrt {5}}-1}{2}}\\\beta ={\dfrac {{\sqrt {5}}+1}{2}}\end{cases}}}
和
b
n
=
a
n
β
n
{\displaystyle b_{n}={\frac {a_{n}}{\beta ^{n}}}}
可得
a
n
=
5
5
⋅
[
(
1
+
5
2
)
n
−
(
1
−
5
2
)
n
]
{\displaystyle a_{n}={\frac {\sqrt {5}}{5}}\cdot \left[\left({\frac {1+{\sqrt {5}}}{2}}\right)^{n}-\left({\frac {1-{\sqrt {5}}}{2}}\right)^{n}\right]}
得出
a
n
{\displaystyle {a_{n}}}
表達式，稱比內公式(Binet's Formula)
a
n
=
5
5
⋅
[
(
1
+
5
2
)
n
−
(
1
−
5
2
)
n
]
{\displaystyle a_{n}={\frac {\sqrt {5}}{5}}\cdot \left[\left({\frac {1+{\sqrt {5}}}{2}}\right)^{n}-\left({\frac {1-{\sqrt {5}}}{2}}\right)^{n}\right]}
用數學歸納法證明表達式
编辑
證明
F
n
=
1
5
[
φ
n
−
(
1
−
φ
)
n
]
{\displaystyle F_{n}={\frac {1}{\sqrt {5}}}[\varphi ^{n}-(1-\varphi )^{n}]}
，其中
φ
{\displaystyle \varphi }
為黃金比例
1
+
5
2
{\displaystyle {\frac {1+{\sqrt {5}}}{2}}}
，
n
{\displaystyle n}
為任意整數
若
n
{\displaystyle n}
為非負整數
當
n
=
0
{\displaystyle n=0}
時，
1
5
[
φ
0
−
(
1
−
φ
)
0
]
=
1
5
[
1
−
1
]
=
0
=
F
0
{\displaystyle {\frac {1}{\sqrt {5}}}[\varphi ^{0}-(1-\varphi )^{0}]={\frac {1}{\sqrt {5}}}[1-1]=0=F_{0}}
，成立
當
n
=
1
{\displaystyle n=1}
時，
1
5
[
φ
1
−
(
1
−
φ
)
1
]
=
1
5
[
φ
−
1
+
φ
]
=
1
5
[
2
φ
−
1
]
=
1
5
×
5
=
1
=
F
1
{\displaystyle {\frac {1}{\sqrt {5}}}[\varphi ^{1}-(1-\varphi )^{1}]={\frac {1}{\sqrt {5}}}[\varphi -1+\varphi ]={\frac {1}{\sqrt {5}}}[2\varphi -1]={\frac {1}{\sqrt {5}}}\times {\sqrt {5}}=1=F_{1}}
，成立
設當
n
=
k
{\displaystyle n=k}
及
n
=
k
+
1
{\displaystyle n=k+1}
時皆成立，即
F
k
=
1
5
[
φ
k
−
(
1
−
φ
)
k
]
{\displaystyle F_{k}={\frac {1}{\sqrt {5}}}[\varphi ^{k}-(1-\varphi )^{k}]}
且
F
k
+
1
=
1
5
[
φ
k
+
1
−
(
1
−
φ
)
k
+
1
]
{\displaystyle F_{k+1}={\frac {1}{\sqrt {5}}}[\varphi ^{k+1}-(1-\varphi )^{k+1}]}
當
n
=
k
+
2
{\displaystyle n=k+2}
時
F
k
+
2
=
F
k
+
1
+
F
k
=
1
5
[
φ
k
+
1
−
(
1
−
φ
)
k
+
1
]
+
1
5
[
φ
k
−
(
1
−
φ
)
k
]
=
1
5
[
φ
k
+
1
+
φ
k
−
(
1
−
φ
)
k
+
1
−
(
1
−
φ
)
k
]
=
1
5
{
φ
k
(
φ
+
1
)
−
(
1
−
φ
)
k
[
(
1
−
φ
)
+
1
]
}
=
1
5
{
φ
k
(
φ
2
)
−
(
1
−
φ
)
k
[
(
1
−
φ
)
2
]
}
=
1
5
{
φ
k
+
2
−
(
1
−
φ
)
k
+
2
}
{\displaystyle {\begin{aligned}F_{k+2}&=F_{k+1}+F_{k}\\&={\frac {1}{\sqrt {5}}}[\varphi ^{k+1}-(1-\varphi )^{k+1}]+{\frac {1}{\sqrt {5}}}[\varphi ^{k}-(1-\varphi )^{k}]\\&={\frac {1}{\sqrt {5}}}[\varphi ^{k+1}+\varphi ^{k}-(1-\varphi )^{k+1}-(1-\varphi )^{k}]\\&={\frac {1}{\sqrt {5}}}\left\{\varphi ^{k}({\color {brown}\varphi +1})-(1-\varphi )^{k}[{\color {green}(1-\varphi )+1}]\right\}\\&={\frac {1}{\sqrt {5}}}\left\{\varphi ^{k}({\color {brown}\varphi ^{2}})-(1-\varphi )^{k}[{\color {green}(1-\varphi )^{2}}]\right\}\\&={\frac {1}{\sqrt {5}}}\left\{\varphi ^{k+2}-(1-\varphi )^{k+2}\right\}\\\end{aligned}}}
亦成立
若
n
{\displaystyle n}
為非正整數
當
n
=
0
{\displaystyle n=0}
時，成立
當
n
=
−
1
{\displaystyle n=-1}
時，
1
5
[
φ
−
1
−
(
1
−
φ
)
−
1
]
=
1
5
[
(
φ
−
1
)
−
(
−
φ
)
]
=
1
5
[
2
φ
−
1
]
=
1
5
×
5
=
1
=
F
−
1
{\displaystyle {\frac {1}{\sqrt {5}}}[{\color {brown}\varphi ^{-1}}-{\color {green}(1-\varphi )^{-1}}]={\frac {1}{\sqrt {5}}}[({\color {brown}\varphi -1})-({\color {green}-\varphi })]={\frac {1}{\sqrt {5}}}[2\varphi -1]={\frac {1}{\sqrt {5}}}\times {\sqrt {5}}=1=F_{-1}}
，成立
設當
n
=
−
k
{\displaystyle n=-k}
及
n
=
−
k
−
1
{\displaystyle n=-k-1}
時皆成立，即
F
−
k
=
1
5
[
φ
−
k
−
(
1
−
φ
)
−
k
]
{\displaystyle F_{-k}={\frac {1}{\sqrt {5}}}[\varphi ^{-k}-(1-\varphi )^{-k}]}
且
F
−
k
−
1
=
1
5
[
φ
−
k
−
1
−
(
1
−
φ
)
−
k
−
1
]
{\displaystyle F_{-k-1}={\frac {1}{\sqrt {5}}}[\varphi ^{-k-1}-(1-\varphi )^{-k-1}]}
當
n
=
−
k
−
2
{\displaystyle n=-k-2}
時
F
−
k
−
2
=
F
−
k
−
F
−
k
−
1
=
1
5
[
φ
−
k
−
(
1
−
φ
)
−
k
]
−
1
5
[
φ
−
k
−
1
−
(
1
−
φ
)
−
k
−
1
]
=
1
5
[
φ
−
k
−
φ
−
k
−
1
−
(
1
−
φ
)
−
k
+
(
1
−
φ
)
−
k
−
1
]
=
1
5
{
φ
−
k
−
1
(
φ
−
1
)
−
(
1
−
φ
)
−
k
−
1
[
(
1
−
φ
)
−
1
]
}
=
1
5
{
φ
−
k
−
1
(
φ
−
1
)
−
(
1
−
φ
)
−
k
−
1
[
(
1
−
φ
)
−
1
]
}
=
1
5
{
φ
−
k
−
2
−
(
1
−
φ
)
−
k
−
2
}
{\displaystyle {\begin{aligned}F_{-k-2}&=F_{-k}-F_{-k-1}\\&={\frac {1}{\sqrt {5}}}[\varphi ^{-k}-(1-\varphi )^{-k}]-{\frac {1}{\sqrt {5}}}[\varphi ^{-k-1}-(1-\varphi )^{-k-1}]\\&={\frac {1}{\sqrt {5}}}[\varphi ^{-k}-\varphi ^{-k-1}-(1-\varphi )^{-k}+(1-\varphi )^{-k-1}]\\&={\frac {1}{\sqrt {5}}}\left\{\varphi ^{-k-1}({\color {brown}\varphi -1})-(1-\varphi )^{-k-1}[{\color {green}(1-\varphi )-1}]\right\}\\&={\frac {1}{\sqrt {5}}}\left\{\varphi ^{-k-1}({\color {brown}\varphi ^{-1}})-(1-\varphi )^{-k-1}[{\color {green}(1-\varphi )^{-1}}]\right\}\\&={\frac {1}{\sqrt {5}}}\left\{\varphi ^{-k-2}-(1-\varphi )^{-k-2}\right\}\\\end{aligned}}}
亦成立
因此，根據數學歸納法原理，此表達式對於任意整數
n
{\displaystyle n}
皆成立
線性代數解法
编辑
(
F
n
+
2
F
n
+
1
)
=
(
1
1
1
0
)
⋅
(
F
n
+
1
F
n
)
{\displaystyle {\begin{pmatrix}F_{n+2}\\F_{n+1}\end{pmatrix}}={\begin{pmatrix}1&1\\1&0\end{pmatrix}}\cdot {\begin{pmatrix}F_{n+1}\\F_{n}\end{pmatrix}}}
(
F
n
+
2
F
n
+
1
F
n
+
1
F
n
)
=
(
1
1
1
0
)
n
+
1
{\displaystyle {\begin{pmatrix}F_{n+2}&F_{n+1}\\F_{n+1}&F_{n}\end{pmatrix}}={\begin{pmatrix}1&1\\1&0\end{pmatrix}}^{n+1}}
稱為「斐波那契Q矩陣」(Fibonacci Q-Matrix)[1]
構建一個矩陣方程
编辑
設
J
n
{\displaystyle J_{n}}
為第
n
{\displaystyle n}
個月有生育能力的兔子數量，
A
n
{\displaystyle A_{n}}
為這一月份的兔子數量。
(
J
n
+
1
A
n
+
1
)
=
(
0
1
1
1
)
⋅
(
J
n
A
n
)
,
{\displaystyle {J_{n+1} \choose A_{n+1}}={\begin{pmatrix}0&1\\1&1\end{pmatrix}}\cdot {J_{n} \choose A_{n}},}
上式表達了兩個月之間，兔子數目之間的關係。而要求的是，
A
n
+
1
{\displaystyle A_{n+1}}
的表達式。
求矩陣的特徵值：λ
编辑
根据特征值的计算公式，我们需要算出来
|
−
λ
1
1
1
−
λ
|
=
0
{\displaystyle {\begin{vmatrix}-\lambda &1\\1&1-\lambda \\\end{vmatrix}}=0}
所对应的解。
展开行列式有：
−
λ
(
1
−
λ
)
−
1
×
1
=
λ
2
−
λ
−
1
{\displaystyle -\lambda (1-\lambda )-1\times 1=\lambda ^{2}-\lambda -1}
。
故當行列式的值為 0，解得
λ
1
=
1
2
(
1
+
5
)
{\displaystyle \lambda _{1}={\frac {1}{2}}(1+{\sqrt {5}})}
或
λ
2
=
1
2
(
1
−
5
)
{\displaystyle \lambda _{2}={\frac {1}{2}}(1-{\sqrt {5}})}
。
特徵向量
编辑
將兩個特徵值代入
(
(
0
1
1
1
)
−
λ
⋅
E
)
⋅
x
→
=
0
{\displaystyle \left({\begin{pmatrix}0&1\\1&1\end{pmatrix}}-\lambda \cdot E\right)\cdot {\vec {x}}=0}
求特徵向量
x
→
{\displaystyle {\vec {x}}}
得
x
→
1
{\displaystyle {\vec {x}}_{1}}
=
(
1
1
2
(
1
+
5
)
)
{\displaystyle {\begin{pmatrix}1\\{\frac {1}{2}}(1+{\sqrt {5}})\end{pmatrix}}}
x
→
2
{\displaystyle {\vec {x}}_{2}}
=
(
1
1
2
(
1
−
5
)
)
{\displaystyle {\begin{pmatrix}1\\{\frac {1}{2}}(1-{\sqrt {5}})\end{pmatrix}}}
分解首向量
编辑
第一個月的情況是兔子一對，新生0對。
(
J
1
A
1
)
=
(
0
1
)
{\displaystyle {J_{1} \choose A_{1}}={\begin{pmatrix}0\\1\end{pmatrix}}}
將它分解為用特徵向量表示。
(
0
1
)
=
1
5
⋅
(
1
1
2
(
1
+
5
)
)
−
1
5
⋅
(
1
1
2
(
1
−
5
)
)
{\displaystyle {\begin{pmatrix}0\\1\end{pmatrix}}={\frac {1}{\sqrt {5}}}\cdot {\begin{pmatrix}1\\{\frac {1}{2}}(1+{\sqrt {5}})\end{pmatrix}}-{\frac {1}{\sqrt {5}}}\cdot {\begin{pmatrix}1\\{\frac {1}{2}}(1-{\sqrt {5}})\end{pmatrix}}}
（4）
用數學歸納法證明
编辑
從
(
J
n
+
1
A
n
+
1
)
=
(
0
1
1
1
)
⋅
(
J
n
A
n
)
{\displaystyle {J_{n+1} \choose A_{n+1}}={\begin{pmatrix}0&1\\1&1\end{pmatrix}}\cdot {J_{n} \choose A_{n}}}
=
λ
⋅
(
J
n
A
n
)
{\displaystyle \lambda \cdot {J_{n} \choose A_{n}}}
可得到
(
J
n
+
1
A
n
+
1
)
=
(
0
1
1
1
)
n
⋅
(
J
1
A
1
)
=
λ
n
⋅
(
J
1
A
1
)
{\displaystyle {J_{n+1} \choose A_{n+1}}={\begin{pmatrix}0&1\\1&1\end{pmatrix}}^{n}\cdot {J_{1} \choose A_{1}}=\lambda ^{n}\cdot {J_{1} \choose A_{1}}}
（5）
化簡矩陣方程
编辑
將（4） 代入 （5）
(
J
n
+
1
A
n
+
1
)
=
λ
n
⋅
[
1
5
⋅
(
1
1
2
(
1
+
5
)
)
−
1
5
⋅
(
1
1
2
(
1
−
5
)
)
]
{\displaystyle {J_{n+1} \choose A_{n+1}}=\lambda ^{n}\cdot \left[{\frac {1}{\sqrt {5}}}\cdot {\begin{pmatrix}1\\{\frac {1}{2}}(1+{\sqrt {5}})\end{pmatrix}}-{\frac {1}{\sqrt {5}}}\cdot {\begin{pmatrix}1\\{\frac {1}{2}}(1-{\sqrt {5}})\end{pmatrix}}\right]}
根據3
(
J
n
+
1
A
n
+
1
)
=
1
5
⋅
λ
1
n
⋅
(
1
1
2
(
1
+
5
)
)
−
1
5
⋅
λ
2
n
⋅
(
1
1
2
(
1
−
5
)
)
{\displaystyle {J_{n+1} \choose A_{n+1}}={\frac {1}{\sqrt {5}}}\cdot \lambda _{1}^{n}\cdot {\begin{pmatrix}1\\{\frac {1}{2}}(1+{\sqrt {5}})\end{pmatrix}}-{\frac {1}{\sqrt {5}}}\cdot \lambda _{2}^{n}\cdot {\begin{pmatrix}1\\{\frac {1}{2}}(1-{\sqrt {5}})\end{pmatrix}}}
求A的表達式
编辑
現在在6的基礎上，可以很快求出
A
n
+
1
{\displaystyle A_{n+1}}
的表達式，將兩個特徵值代入6中
A
n
+
1
=
1
5
⋅
λ
1
n
+
1
−
1
5
⋅
λ
2
n
+
1
{\displaystyle A_{n+1}={\frac {1}{\sqrt {5}}}\cdot \lambda _{1}^{n+1}-{\frac {1}{\sqrt {5}}}\cdot \lambda _{2}^{n+1}}
A
n
+
1
=
1
5
⋅
(
λ
1
n
+
1
−
λ
2
n
+
1
)
{\displaystyle A_{n+1}={\frac {1}{\sqrt {5}}}\cdot (\lambda _{1}^{n+1}-\lambda _{2}^{n+1})}
A
n
+
1
=
1
5
⋅
{
[
1
2
(
1
+
5
)
]
n
+
1
−
[
1
2
(
1
−
5
)
]
n
+
1
}
{\displaystyle A_{n+1}={\frac {1}{\sqrt {5}}}\cdot \left\{\left[{\frac {1}{2}}\left(1+{\sqrt {5}}\right)\right]^{n+1}-\left[{\frac {1}{2}}(1-{\sqrt {5}})\right]^{n+1}\right\}}
（7）
（7）即為
A
n
+
1
{\displaystyle A_{n+1}}
的表達式
數論解法
编辑
實際上，如果將斐波那契數列的通項公式寫成
a
n
−
a
n
−
1
−
a
n
−
2
=
0
{\displaystyle a_{n}-a_{n-1}-a_{n-2}=0}
，即可利用解二階線性齊次遞迴關係式的方法，寫出其特徵多項式
λ
2
−
λ
−
1
=
0
{\displaystyle \lambda ^{2}-\lambda -1=0}
（該式和表達斐波那契數列的矩陣的特徵多項式一致），然後解出
λ
1
{\displaystyle \lambda _{1}}
=
1
2
(
1
+
5
)
{\displaystyle {\frac {1}{2}}(1+{\sqrt {5}})}
，
λ
2
{\displaystyle \lambda _{2}}
=
1
2
(
1
−
5
)
{\displaystyle {\frac {1}{2}}(1-{\sqrt {5}})}
，即有
a
n
=
c
1
λ
1
n
+
c
2
λ
2
n
{\displaystyle a_{n}=c_{1}\lambda _{1}^{n}+c_{2}\lambda _{2}^{n}}
，其中
c
1
,
c
2
{\displaystyle c_{1},c_{2}}
为常数。我们知道
a
0
=
0
,
a
1
=
1
{\displaystyle a_{0}=0,a_{1}=1}
，因此
{
c
1
+
c
2
=
0
c
1
(
1
+
5
)
2
+
c
2
(
1
−
5
)
2
=
1
{\displaystyle {\begin{cases}c_{1}+c_{2}=0\\{\frac {c_{1}(1+{\sqrt {5}})}{2}}+{\frac {c_{2}(1-{\sqrt {5}})}{2}}=1\end{cases}}}
，解得
c
1
=
1
5
,
c
2
=
−
1
5
{\displaystyle c_{1}={\frac {1}{\sqrt {5}}},c_{2}=-{\frac {1}{\sqrt {5}}}}
。
組合數解法
编辑
F
n
=
∑
i
=
0
∞
(
n
−
i
i
)
{\displaystyle F_{n}=\sum _{i=0}^{\infty }{\binom {n-i}{i}}}
[2]
F
n
−
1
+
F
n
=
∑
i
=
0
∞
(
n
−
1
−
i
i
)
+
∑
i
=
0
∞
(
n
−
i
i
)
=
1
+
∑
i
=
1
∞
(
n
−
i
i
−
1
)
+
∑
i
=
1
∞
(
n
−
i
i
)
=
1
+
∑
i
=
1
∞
(
n
+
1
−
i
i
)
=
∑
i
=
0
∞
(
n
+
1
−
i
i
)
=
F
n
+
1
{\displaystyle F_{n-1}+F_{n}=\sum _{i=0}^{\infty }{\binom {n-1-i}{i}}+\sum _{i=0}^{\infty }{\binom {n-i}{i}}=1+\sum _{i=1}^{\infty }{\binom {n-i}{i-1}}+\sum _{i=1}^{\infty }{\binom {n-i}{i}}=1+\sum _{i=1}^{\infty }{\binom {n+1-i}{i}}=\sum _{i=0}^{\infty }{\binom {n+1-i}{i}}=F_{n+1}}
黃金比例恆等式解法
编辑
設
φ
{\displaystyle \varphi }
為黃金比例
1
+
5
2
{\displaystyle {\frac {1+{\sqrt {5}}}{2}}}
，則有恆等式
φ
n
=
F
n
−
1
+
φ
F
n
{\displaystyle \varphi ^{n}=F_{n-1}+\varphi F_{n}}
與
(
1
−
φ
)
n
=
F
n
+
1
−
φ
F
n
{\displaystyle (1-\varphi )^{n}=F_{n+1}-\varphi F_{n}}
，其中
n
{\displaystyle n}
為任意整數[註 1]，則
φ
n
−
(
1
−
φ
)
n
=
(
F
n
−
1
+
φ
F
n
)
−
(
F
n
+
1
−
φ
F
n
)
=
(
F
n
−
1
−
F
n
+
1
)
+
2
φ
F
n
=
−
F
n
+
2
φ
F
n
=
F
n
(
2
φ
−
1
)
=
F
n
×
5
{\displaystyle {\begin{aligned}\varphi ^{n}-(1-\varphi )^{n}&=(F_{n-1}+\varphi F_{n})-(F_{n+1}-\varphi F_{n})\\&=(F_{n-1}-F_{n+1})+2\varphi F_{n}\\&=-F_{n}+2\varphi F_{n}\\&=F_{n}(2\varphi -1)\\&=F_{n}\times {\sqrt {5}}\\\end{aligned}}}
因此得到
F
n
{\displaystyle F_{n}}
的一般式：
F
n
=
1
5
[
φ
n
−
(
1
−
φ
)
n
]
=
1
5
[
(
1
+
5
2
)
n
−
(
1
−
5
2
)
n
]
{\displaystyle {\begin{aligned}F_{n}&={\frac {1}{\sqrt {5}}}[\varphi ^{n}-(1-\varphi )^{n}]\\&={\frac {1}{\sqrt {5}}}\left[({\frac {1+{\sqrt {5}}}{2}})^{n}-({\frac {1-{\sqrt {5}}}{2}})^{n}\right]\\\end{aligned}}}
此一般式對任意整數
n
{\displaystyle n}
成立
近似值
编辑
當
n
{\displaystyle n}
為足夠大的正整數時，则
F
n
≈
1
5
φ
n
=
1
5
⋅
[
1
2
(
1
+
5
)
]
n
≈
0.4472135955
⋅
1.61803398875
n
{\displaystyle F_{n}\approx {\frac {1}{\sqrt {5}}}\varphi ^{n}={\frac {1}{\sqrt {5}}}\cdot \left[{\frac {1}{2}}\left(1+{\sqrt {5}}\right)\right]^{n}\approx 0.4472135955\cdot 1.61803398875^{n}}
F
−
n
≈
−
1
5
(
1
−
φ
)
−
n
=
−
1
5
⋅
[
1
2
(
1
−
5
)
]
−
n
≈
−
0.4472135955
⋅
(
−
0.61803398875
)
−
n
{\displaystyle F_{-n}\approx -{\frac {1}{\sqrt {5}}}(1-\varphi )^{-n}=-{\frac {1}{\sqrt {5}}}\cdot \left[{\frac {1}{2}}\left(1-{\sqrt {5}}\right)\right]^{-n}\approx -0.4472135955\cdot (-0.61803398875)^{-n}}
用計算機求解
编辑
可通過編程觀察斐波那契數列。分為兩類問題，一種已知數列中的某一項，求序數。第二種是已知序數，求該項的值。
可通過遞歸遞推的算法解決此兩個問題。
事實上當
n
{\displaystyle n}
相當巨大的時候，O（n）的遞推/遞歸非常慢……這時候要用到矩陣快速幂這一技巧，可以使遞迴加速到O(logn)。