The explicit formula for the terms of the Fibonacci sequence, $F_n=\dfrac{\left(\dfrac{1+\sqrt{5}}{2}\right)^n-\left(\dfrac{1-\sqrt{5}}{2}\right)^n}{\sqrt{5}}$. has been named in honor of the eighteenth century French mathematician Jacques Binet, although he was not the first to use it. Typically, the formula is proven as a special case of a more general study of sequences in number theory. However, we shall relate the formula to a geometric series.

**Lemma 1:** Let $F_n$ represent the $n$th term of the Fibonacci sequence, and define $E_n = F_n - \left(\dfrac{1+\sqrt{5}}{2}\right)F_{n-1}$. Then $E_n = \left(\dfrac{1-\sqrt{5}}{2}\right)^{n-1}$.

**Proof.**

Let $F_n$ represent the $n$th term of the Fibonacci sequence. Define $E_n = F_n - \left(\dfrac{1+\sqrt{5}}{2}\right)F_{n-1}$. Then we have the following implications: | We must begin with the hypothesis of the lemma. |

$E_n = (F_{n-1}+F_{n-2}) - \left(\dfrac{1+\sqrt{5}}{2}\right)F_{n-1}$ | We apply a number of algebraic properties, beginning with substitution of the recursive formula of $F_n$. |

$E_n = \left(1-\dfrac{1+\sqrt{5}}{2}\right)F_{n-1}+F_{n-2}$ | We collect like terms ... |

$E_n = \left(\dfrac{1-\sqrt{5}}{2}\right) \left[F_{n-1}+\left(\dfrac{2}{1-\sqrt{5}}\right)F_{n-2}\right]$ | ... and simplify the expression in parentheses. Inside the brackets, we introduce its reciprocal as a coefficient of $F_{n-2}$ by "factoring". |

$E_n = \left(\dfrac{1-\sqrt{5}}{2}\right)\left[F_{n-1}-\left(\dfrac{1+\sqrt{5}}{2}\right)F_{n-2}\right]$ | Here we rationalize the denominator. |

But the result in brackets is the error term for $n-1$. That is: $E_n = \left(\dfrac{1-\sqrt{5}}{2}\right)E_{n-1}$ | We now have a recursive formula for $E_n$. |

By repeated use of this result, we can obtain: $E_n = \left(\dfrac{1-\sqrt{5}}{2}\right)^{n-2}E_2$ | In other words, $E_{n-1}$ can be written in terms of $E_{n-2}$, which can be written in terms of $E_{n-3}$, et cetera, until we reach $E_2$. |

And applying the definition of $E_n$ to $E_2$ gives: $E_n = \left(\dfrac{1-\sqrt{5}}{2}\right)^{n-2} \left[F_2-\left(\dfrac{1+\sqrt{5}}{2}\right) F_1\right]$ $E_n = \left(\dfrac{1-\sqrt{5}}{2}\right)^{n-2}\left[1-\left(\dfrac{1+\sqrt{5}}{2}\right)\right]$ |
Then we substitute values for $F_1$ and $F_2$. |

$E_n = \left(\dfrac{1-\sqrt{5}}{2}\right)^{n-1}$ | And lastly, we simplify the result. |

It is not obvious that there should be a connection between Fibonacci sequences and geometric series. Yet once this has been achieved, we will be able to use formulas for geometric series to write our proof of Binet's Formula.

**Lemma 2:** Each term of the Fibonacci sequence is the sum of a finite geometric series with first term $\left(\dfrac{1+\sqrt{5}}{2}\right)^{n-1}$ and ratio $\dfrac{1-\sqrt{5}}{1+\sqrt{5}}$.

**Proof.**

Let $F_n$ represent the $n$th term of the Fibonacci sequence, and define $E_n = F_n - \left(\dfrac{1+\sqrt{5}}{2}\right)F_{n-1}$. | Since the lemma implies a statement about the Fibonacci sequence, we begin by defining $F_n$. In order to use the result of the first lemma, we also define $E_n$. |

Then $F_n=\left(\dfrac{1+\sqrt{5}}{2}\right)F_{n-1} + E_n$, so $F_n=\left(\dfrac{1+\sqrt{5}}{2}\right)F_{n-1} + \left(\dfrac{1-\sqrt{5}}{2}\right)^{n-1}$ | We solve the definition of $E_n$ for the value of $F_n$, then substitute the result of the first lemma. |

Since we have written $F_n$ in terms of $F_{n-1}$, we can now write $F_{n-1}$ in terms of $F_{n-2}$, and so on. We get: $F_n=\left(\dfrac{1+\sqrt{5}}{2}\right)\left[\left(\dfrac{1+\sqrt{5}}{2}\right)F_{n-2}+\left(\dfrac{1-\sqrt{5}}{2}\right)^{n-2}\right]+\left(\dfrac{1-\sqrt{5}}{2}\right)^{n-1}$ |
We can now use this result recursively, substituting the previous equation into itself, ... |

$F_n=\left(\dfrac{1+\sqrt{5}}{2}\right)\left[\left(\dfrac{1+\sqrt{5}}{2}\right)\left[\left(\dfrac{1+\sqrt{5}}{2}\right)F_{n-3}\right.\right.$ $\qquad\qquad\qquad\qquad \left.\left.+\left(\dfrac{1-\sqrt{5}}{2}\right)^{n-3}\right]+\left(\dfrac{1-\sqrt{5}}{2}\right)^{n-2}\right]+\left(\dfrac{1-\sqrt{5}}{2}\right)^{n-1}$ |
... and again into itself. We begin to recognize a pattern developing ... |

After $n-2$ substitutions, and recalling $F_1=1$, we have: $F_n=\left(\dfrac{1+\sqrt{5}}{2}\right)^{n-1} + \left(\dfrac{1+\sqrt{5}}{2}\right)^{n-2}\left(\dfrac{1-\sqrt{5}}{2}\right) + \ldots$ $\qquad\qquad\qquad\qquad + \left(\dfrac{1+\sqrt{5}}{2}\right)\left(\dfrac{1-\sqrt{5}}{2}\right)^{n-2} + \left(\dfrac{1-\sqrt{5}}{2}\right)^{n-1}$ |
... and here are the terms that result. |

But the $n$ terms on the right hand side of the equation form a geometric sequence with ratio $\dfrac{1-\sqrt{5}}{1+\sqrt{5}}$. Since the right hand side has the proper first term, and $F_n$ is the sum of those terms, we have shown that each term of $F_n$ is the sum of a finite geometric series with the characteristics claimed in the statement of the lemma. | Note that the ratio of consecutive terms is $\dfrac{\left(\frac{1+\sqrt{5}}{2}\right)^{n-k} \left(\frac{1-\sqrt{5}}{2}\right)^{k-1}} {\left(\frac{1+\sqrt{5}}{2}\right)^{n-k+1}\left(\frac{1-\sqrt{5}}{2}\right)^{k-2}} = \dfrac{1-\sqrt{5}}{1+\sqrt{5}}$. |

The first lemma was used to prove the second lemma, and now we can use the second lemma to establish Binet's Formula.

**Theorem.** If $F_n$ is the $n$th term of a Fibonacci sequence, then $F_n = \dfrac{\left(\dfrac{1+\sqrt{5}}{2}\right)^n-\left(\dfrac{1-\sqrt{5}}{2}\right)^n}{\sqrt{5}}$.

**Proof.**

Let $F_n$ be the $n$th term of a Fibonacci sequence. Then by the second lemma, $F_n$ is the sum of a finite geometric series with first term $\left(\dfrac{1+\sqrt{5}}{2}\right)^{n-1}$ and ratio $\dfrac{1-\sqrt{5}}{1+\sqrt{5}}$. | We begin with the hypothesis of the theorem, and immediately use the result of the second lemma. |

Since the formula for the sum of a finite geometric sequence is $\sum\limits_{k=1}^{n} a_k = \dfrac{a_1 (1-r^n)}{1-r}$ we have $F_n = \dfrac{\left(\frac{1+\sqrt{5}}{2}\right)^{n-1}\left[1-\left(\frac{1-\sqrt{5}}{1+\sqrt{5}}\right)^n\right]}{\left(1-\frac{1-\sqrt{5}}{1+\sqrt{5}}\right)}$ | Using the formula for the sum of a finite geometric sequence, we substitute. |

Simplifying this result, we obtain: $F_n=\dfrac{\left(\frac{1+\sqrt{5}}{2}\right)^{n-1} - \left(\frac{1-\sqrt{5}}{1+\sqrt{5}}\right)\left(\frac{1-\sqrt{5}}{2}\right)^{n-1}}{\left(\frac{2\sqrt{5}}{1+\sqrt{5}}\right)}$ |
We distribute in the numerator, and combine terms in the denominator. |

$F_n=\left(\dfrac{1+\sqrt{5}}{2\sqrt{5}}\right)\left[\left(\dfrac{1+\sqrt{5}}{2}\right)^{n-1} - \left(\dfrac{1-\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{1-\sqrt{5}}{2}\right)^{n-1}\right]$ | We then multiply the numerator by the reciprocal of the denominator. |

$F_n=\dfrac{\left(\dfrac{1+\sqrt{5}}{2}\right)^n-\left(\dfrac{1-\sqrt{5}}{2}\right)^n}{\sqrt{5}}$ | After distributing, Binet's Formula is obtained. |