Originally Posted by

**hollywood** These are essentially the same formula. If you add $\displaystyle F_{n+1}$ to both sides of the first formula, you get $\displaystyle F_{n}+F_{n+1}=F_{n+2}-F_{n+1}+F_{n+1}=F_{n+2}$ so $\displaystyle F_{n+2}=F_{n+1}+F_{n}$, which is the second formula with n replaced by by n-1 (i.e. all three subscripts decreased by 1).

But let's take your two formulas and do some algebra:

$\displaystyle F_{n}=F_{n+2}-F_{n+1}$

$\displaystyle F_{n+3}=F_{n+2}+F_{n+1}$

adding $\displaystyle F_{n+1}$ to the first gives:

$\displaystyle F_{n}+F_{n+1}=F_{n+2}$

and substituting into the second gives:

$\displaystyle F_{n+3}=F_{n}+F_{n+1}+F_{n+1}=F_{n}+2F_{n+1}$

and subtracting $\displaystyle F_{n}$ from both sides gives:

$\displaystyle F_{n+3}-F_{n}=2F_{n+1}$.

That's the second formula you were looking for.

Now, if you take:

$\displaystyle F_{n+3}=F_{n+2}+F_{n+1}$

and instead of substituting for $\displaystyle F_{n+2}$, you substitute for $\displaystyle F_{n+1}$ using $\displaystyle F_{n+1}=F_{n+2}-F_{n}$, you get:

$\displaystyle F_{n+3}=F_{n+2}+F_{n+2}-F_{n}$

$\displaystyle F_{n+3}+F_{n}=2F_{n+2}$, which is the first formula you were looking for.

- Hollywood