The Explicit Formula for Fibonacci Sequence

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.

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