Finding a formula for the first n Fibonacci numbers

The first two Fibonacci numbers (\(f_1\) and \(f_2\)) are both 1, while \(f_n = f_{n-1} + f_{n-2}\) for \(n \ge 3\). The first twelve terms are:

\begin{align} &1, &&1, &&2, &&3 \\ &5, &&8, &&13, &&21, \\ &34, &&55, &&89, &&144 \end{align}

The sum of the first \(n\) Fibonacci numbers are shown below:

\begin{align} &1, &&2, &&4, &&7 \\ &12, &&20, &&33, &&54, \\ &88, &&143, &&232, &&376 \end{align}

We are trying to find a formula for the sum of the first \(n\) Fibonacci numbers:

$$ \sum^n_{k=1} f_k $$

Since \(f_n = f_{n-1} + f_{n-2}\), we can rewrite this as \(f_{n+2} = f_{n+1} + f_n\), or \(f_n = f_{n+2} - f_{n+1}\). This means:

\begin{align} \sum^n_{k=1} f_k &= \sum^n_{k=1} (f_{k+2} - f_{k+1}) \\ &= \sum^n_{k=1} f_{k+2} - \sum^n_{k=1} f_{k+1} \end{align}

Expanding:

$$ \sum^n_{k=1} f_k = (f_3 + f_4 + f_5 + \cdots + f_{n+1} + f_{n+2}) - (f_2 + f_3 + f_4 + f_5 + \cdots + f_{n+1}) $$

Most of the terms cancel out:

$$ \sum^n_{k=1} f_k = f_{n+2} - f_2 $$

This shows that the sum of the first \(n\) Fibonacci numbers is just the value of \(f_{n+2} - f_2\), or \(f_{n+2} - 1\).

If the first two Fibonacci numbers were \(f_1=7\) and \(f_2=4\), then the rest of the terms would be 11, 15, 26, 41, 67, 108, 175, 283, ..., and the formula for the sum would be \([f_{n+2} - 4]\):

\begin{gather} 7, 11, 22, 37, \\ 63, 104, 171, 279 \end{gather}

Finding A Formula For The First n Fibonacci Numbers