Taylor Series and equation I have this equation:
$$960 - 84.60 \cdot \frac{1-(1+i)^{-12}}{i} = 0$$
I simplify $( 1+i)^{-12}$ with a Taylor series $( 1 + x)^a$.
but I obtain $i = 0.087201167$ but the real result should be $i = 0.00753$ (approximately). 
P.S.: My solution
$$( 1 + i)^{-12} = 1 - 12i + 78i^2 + 364i^3 $$
Then:
$$1 -(1+i)^{-12} = 12i - 78i^2 + 364i^3$$
Then I collect i and I simplify so I obtain:
$$960 - 84.60 \cdot (12 - 78 i + 364 i^2) = 0$$
And I obtain:
$$-55.20 + 6598.80 i - 30794.40i^2 == 0$$
I apply the Quadratic Formula and I obtain:
$$i = 0.087201167` \quad  i = 2.055`$$
 A: Your equation can only be solved numerically.
This means that an iterative method has to be used
starting at an approximate solution.
I would solve it like this:
Since your equation is
$960 - 84.60 * \left(\dfrac{1-(1+i)^{-12}}{i}\right) = 0
$,
I would write it in the form
$1-(1+i)^{-12} = a i$,
where
$a = \dfrac{960}{84.60}\sim 11.34$.
I would then use the first two terms
in the expansion of $(1+i)^{-12}$
to get an initial value for $i$.
Since 
$$(1+i)^{-12}
\sim 1-12i +\frac{(-12)(-13)}{2}i^2 
= 1-12i +78i^2,
$$
$$\dfrac{1-(1+i)^{-12}}{i}
\sim \dfrac{1-(1-12i+78i^2)}{i}
=12-78i
,$$
so the initial approximation would be
$12-78i = 11.34$
or
$i = \frac{.66}{78}
\sim 0.0084
$.
From this initial value for $i$,
I would apply Newton's rule:
if $x$ is a approximate root of
$f(x) = 0$,
a better root is
$x-\dfrac{f(x)}{f'(x)}$.
If $f(i) 
= 1-(1+i)^{-12} - a i
$,
$f'(i)
=12(1+i)^{-13}-a
$,
so the iteration would be
$i 
=i - \dfrac{1-ai-(1+i)^{-12}}{12(1+i)^{-13}-a}
$.
You could also write this,
by multiplying top and bottom by
$(1+i)^{13}$,
as
$i
=i - \dfrac{(1+i)^{13}(1-ai)-(1+i)}{12-a(1+i)^{13}}
$
In either case,
I would leave the iteration in the form
$i = i-...$
so I could see when it converges
by looking at how much $i$
is changed.
