There was a question in a mental math test and it expected me to calculate the first 9 terms of the following Fibonacci sequence (Note that this is how the problem was exactly given.):
The sum of the first nine terms of the Fibonacci sequence 4,7,11,18,29... is
How could I have done this in under 15 seconds? I know that if there are n terms, then $2F_n+F_{n-1}=$ Sum of $n$ Fibonacci numbers. But even this is too tedious for me.
On
Notice that
$$F_n-F_{n-1}=F_{n-2}$$
So
\begin{align*} F_9=F_{11}-\color{magenta}{F_{10}}\\ F_8=\color{magenta}{F_{10}}-\color{red}{F_9}\\ F_7=\color{red}{F_9}-\color{blue}{F_8}\\ F_6=\color{blue}{F_8}-\color{green}{F_7}\\ F_5=\color{green}{F_7}-\color{purple}{F_6}\\ F_4=\color{purple}{F_6}-\color{orange}{F_5}\\ F_3=\color{orange}{F_5}-\color{yellow}{F_4}\\ F_2=\color{yellow}{F_4}-\color{pink}{F_3}\\ F_1=\color{pink}{F_3}-{F_2}\\ \end{align*}
So the sum of all is $F_{11}-F_2$
On
Maybe you know the standard Fibonacci numbers by heart. The others have shown that the sum of the first $9$ terms is the eleventh term minus the second. You can use the linearity of the recurrence to say that the series that begins $4,7$ is four times the series that begins $1,1$ plus three times the series that begins $0,1$, which is the same as the usual series offset downward by $1$. Thus if we use $F_n$ for the standard Fibonacci numbers, your sum is $4F_{11}-4F_2 +3F_{10}-3F_1=4\cdot 89 - 4 \cdot 1 +3 \cdot 55 -3 \cdot 1=514$
Fibonacci sequence follows the recurrence relation: $$F_{n+1}=F_{n}+F_{n-1}$$ And this can be re-arranged as $$F_{n-1}=F_{n+1}-F_{n}$$
Now using this, your sum will become a telescopic series. See if you can complete it now.
UPDATE: As mentioned by MonsieurGalois, it turns out to be $F_{11}-F_2$.
ADDENDUM: In general, it must be kept in mind (for a generalised expression) that the sum of the first $n $ terms of the Fibonacci sequence gives the sum as $F_{n+2}-F_2$.