First, let's write out the recursive formula: \[a_{n+2}=a_{n+1}+a_n\] where \(a_{ 1 }=1,\quad a_2=1\)

Now, the expression will be modified in 2 different, but similar, ways.

**First modification**: \[a_{n+2}-\alpha a_{n+1}=\beta (a_{n+1}-\alpha a_ n)\]
Define a new sequence like this: \[b_n=a_{ n+1 }-\alpha a_{ n }\]
Then, the modified version of the Fibonacci Sequence looks like this: \[b_{n+1}=\beta b_n\]
As you can see, it's telling us that the sequence \(b_n\) is a geometric progression, so we can write the explicit form of \(b_n\) like this: \[b_{ n }=b_{ 1 }\beta ^{n-1}=(a_{1+1}-\alpha a_1)\beta^{n-1}=(a_{2}-\alpha a_1)\beta^{n-1}=(1-\alpha )\beta^{n-1} \]

**Second modification**: \[a_{n+2}-\beta a_{n+1}=\alpha (a_{n+1}-\beta a_ n)\]
Define a new sequence like this: \[c_n=a_{ n+1 }-\beta a_{ n }\]
Then, the modified version of the Fibonacci Sequence looks like this: \[c_{n+1}=\alpha c_n\]
As you can see, it's telling us that the sequence \(c_n\) is a geometric progression, so we can write the explicit form of \(c_n\) like this: \[c_{ n }=c_{ 1 }\alpha ^{n-1}=(a_{1+1}-\beta a_1)\alpha^{n-1}=(a_{2}-\beta a_1)\alpha^{n-1}=(1-\beta )\alpha^{n-1}\]

Now, recall that the original version of the Fibonacci Sequence was \(a_{n+2}=a_{n+1}+a_n\). Rewrite it as \[1a_{n+2}-1a_{n+1}-1a_n=0\]

Our **first modified version of the Fibonacci Sequence** was \(a_{n+2}-\alpha a_{n+1}=\beta (a_{n+1}-\alpha a_ n)\), and our **second modified version of the Fibonacci Sequence** was \(a_{n+2}-\beta a_{n+1}=\alpha (a_{n+1}-\beta a_ n)\)

Both of them can be rewritten as \[1a_{ n+2 }-(\alpha +\beta )a_{n+1}+\alpha \beta a_n=0\]

The above 2 equations have to be the same, so we can say that \[\alpha \beta =-1,\quad \alpha +\beta =1\Longrightarrow \alpha =1-\beta \quad and\quad \beta =1-\alpha \]

Substituting it into the 2 modifications yields...

**First Modification**: \[b_n=(1-\alpha )\beta^{n-1}=\beta \times \beta ^{n-1}=\beta^n\]
Recalling our definition of \(b_n\), \[a_{n+1}-\alpha a_n=\beta ^n\]

**Second Modification**: \[c_n=(1-\beta )\alpha^{n-1}=\alpha \times \alpha ^{n-1}=\alpha^n\]
Recalling our definition of \(c_n\), \[a_{n+1}-\beta a_n=\alpha ^n\]

(We're almost done) Subtracting those 2 equations gives us \[\{ a_{ n+1 }-\alpha a_{ n }\} -\{ a_{ n+1 }-\beta a_{ n }\} =\beta ^{ n }-\alpha ^{ n }\longrightarrow (\beta -\alpha )a_{ n }=\beta ^{ n }-\alpha ^{ n }\\ \therefore a_{ n }=\frac { \beta ^{ n }-\alpha ^{ n } }{ \beta -\alpha } \] which is the explicit form of the Fibonacci Sequence.

If we can find the values of \(\alpha\) and \(\beta\), then we'll be done. With a bit of thinking, this is quite easy.

Remember that \(\alpha \beta =-1,\quad \alpha +\beta =1\) This is telling us that \(\alpha\) and \(\beta\) are the roots of \(x^2-x-1=0\) since the sum is 1 and the product is -1. Solving it yields \[\alpha =\frac { 1-\sqrt { 5 } }{ 2 } ,\quad \beta =\frac { 1+\sqrt { 5 } }{ 2 } \]

Finally, substitute the values of \(\alpha\) and \(\beta\) into the explicit formula.

\[a_{ n }=\frac { \beta ^{ n }-\alpha ^{ n } }{ \beta -\alpha } =\frac { \left( \frac { 1+\sqrt { 5 } }{ 2 } \right) ^{ n }-\left( \frac { 1-\sqrt { 5 } }{ 2 } \right) ^{ n } }{ \frac { 1+\sqrt { 5 } }{ 2 } -\frac { 1-\sqrt { 5 } }{ 2 } } =\frac { \left( \frac { 1+\sqrt { 5 } }{ 2 } \right) ^{ n }-\left( \frac { 1-\sqrt { 5 } }{ 2 } \right) ^{ n } }{ \sqrt { 5 } } \]

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## Comments

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TopNewest@Calvin Lin I learned this method from my math teacher, but is there a much easier way to derive the explicit formula for the Fibonacci Sequence?

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See Linear Recurrence Relations, which was from a note written up by @Daniel Chiu

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한국분이세요? 피보나치 수열은 극한값으로구하고 특성방정식으로 끼워넣으시는게 제일편해요!

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한번 보여주실 수 있으세요? 아직 수열은 배우고 있는 중이라서...

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고등학교 과정은 아니지만 특성방정식이라는게 있어요. 그게 이제 극한의 개념을 약간 이용한건데 수열 배우시면 극한도 혹시 배우셨나요? 그럼 이야기가 좀 편해져서요~^^

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