Calculating $\frac{1.01^5}{1.01^5-1}$ without calculator with good accuracy? Question (may SKIP reading this):

A computer is sold either for $19200$ cash or for $4800$ cash down payment together with five equal monthly installments. If the rate of interest charged is $12\%$ per annum, then the amount of each installment (nearest to a rupee) is?
Options:
A)$2965\qquad$ B)$2990\qquad$ C)$3016\qquad$ D)$2896\qquad$ E)$2880$

Boiling down to calculating:

$$\frac{144(1.01)^5}{1.01^5-1}$$
Where if I were to approximate $1.01^5\sim1+0.01(5)=1.05$, returns $3024$ as answer thus tempting one to select option 'C' which is wrong!

The answer option is Option A (which is itself a bit too off from the more accurate value of $2967$ but makes sense anyways as other options are too far apart from this value).
So, without using a calculator, how to calculate the above expression with greater accuracy?

Note:
Just acknowledging the fact that I received a lot of great answers but sadly could accept only $1$ which turned out to be a very difficult task. 
Finally, unable to select on my own, I went with the one that the community selected (most upvoted).
 A: By using $a^5-b^5=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)$, I would do
\begin{align}
\frac{144\cdot1.01^5}{1.01^5-1} &=
  \frac{144\cdot1.01^5}{(1.01-1)(1.01^4+1.01^3+1.01^2+1.01+1)} = \\
  &= \frac{144\cdot1.01^5}{0.01(1.01^4+1.01^3+1.01^2+1.01+1)} = \\
  &= \frac{14400\cdot1.01^5}{1.01^4+1.01^3+1.01^2+1.01+1} \approx \\
  &\approx \frac{14400\cdot1.05}{1.04+1.03+1.02+1.01+1} = \\
  &= \frac{14400\cdot1.05}{5.1} \approx 2964.71 \\
\end{align}
A: You just need to note that $(1+x)^5$ has a full binomial expansion given by $$
(1+x)^5= 1+5x+10x^2+10x^3+5x^4+x^5
$$
Therefore, if the approximation $(1+x)^5 \approx 1+5x$ does not work, then try $1 + 5x + 10x^2$ instead. This leads to $1.051$ as an approximation. If necessary, going one further down the expansion $1.05101$, which upon substitution will not change the answer much. This is the stipulation of some comments in this thread as well : the answer will stabilize near $2967$ by the second order approximation.

In the absence of division
Suppose, however, you wanted to completely avoid division : your calculator hates division (or you hate it). Then, you need to focus on the function $\frac{x^5}{x^5-1}$ and how it behaves near $1$, because $1.01$ is close to $1$.
We write it as $\frac{x^5}{x^5-1} = 1 + \frac{1}{x^5-1}$ so that we only need to focus on $\frac{1}{x^5-1}$. However, there's a problem : we cannot "Taylor expand" $\frac{1}{x^5-1}$ around the point $1$, because it goes to $+\infty$ as $x \to 1$. However, we can still isolate the "bad" part by finding the rate at which $\frac{1}{x^5-1}$ goes to $+\infty$ as $x$ goes to $1$. Then, the remaining part will admit a Taylor expansion.
To do that, observe that
$$
\frac{1}{x^5-1} = 1+\frac{1}{(x-1)(1+x+x^2+x^3+x^4)} \approx \frac{1}{5(x-1)}
$$
Basically, $\frac{1}{x^5-1}$ behaves "like" $\frac{1}{5(x-1)}$ as $x$ is closed to $1$. We are led to expect that removing the "bad" part $\frac{1}{5(x-1)}$ from $\frac{1}{x^5-1}$ should lead to something that is Taylor expandable around $1$.
You will observe that the approximation $\frac{1}{x^5-1} \approx \frac{1}{5(x-1)}$ is not good enough for the question you're solving (because you'll get $144 \times 21 = 3024$ which is not good enough).
As I said, we need to see if removing the "bad" part $\frac{1}{5(x-1)}$ from $\frac{1}{x^5-1}$ leads to something that is finite around $x \to 1$, so that it can be Taylor expanded if necessary. That's why we consider
$$
\lim_{x \to 1} \frac{1}{x^5-1} - \frac{1}{5(x-1)} = \frac{5x-x^5-4}{5(x-1)(x^5-1)}  
$$
A couple of L'Hospitalizations (quite easy ones, because the product below is easy to expand) later, you land at the quantity $-\frac{2}{5}$. This will tell you that the function $$
g(x) =\frac{1}{x^5-1} - \frac{1}{5(x-1)}
$$
satisfies $\lim_{x \to 1} g(x) = -\frac{2}{5}$. Now, Taylor expanding $g$ around the point $1$ (you can expect $g$ to have a Taylor expansion : that's not a worry)
$$
\frac{1}{x^5-1} - \frac{1}{5(x-1)} \approx - \frac{2}{5} + O((x-1))
$$
near the point $x=1$ (leaving out everything except the constant term in the Taylor expansion of $g$).
Now, trying $x=1.01$ out gives $20.6$, which leads to $144 \times 2.06 = 2964.4$, which is also good enough : and avoids any kind of decimal by-hand division.

Note that we used the following heuristic in the second part : even when a function $h(x)$ is not differentiable at a point $x_0$ , it may happen that we can find $N$ so that $h(x)(x-x_0)^N$ is differentiable at $x_0$. In that case, we can still write down an asymptotic analysis for $h(x)$ near $x_0$. That's basically what we did here for the function $h(x) = \frac{1}{x^5-1}$ at $x_0=1$ with $N=1$, to find the nature of $h$ accurately.
That's an extension of the usual derivative approximation $h(x+a) \approx h(x) + ah'(x)+\ldots$ that we're considering. In complex analysis (where it is frequently used e.g. in the theory of generating functions), this kind of analysis appears when we are working with Laurent series , poles and meromorphic functions, for instance.
A: The thing is, the multiplication of the given expression by $144$ only serves to amplify the error in your computation $144$ times, which makes the answer deviate greatly from the correct answer.
As a general rule of thumb, you should take the Taylor polynomial upto the second degree. If larger numbers, say 2345785, were multiplied to the expression, then I would take the polynomial upto much higher degrees.
So, $1.01^5=1+(0.01)5+\dfrac12(0.01)^2\cdot5\cdot4=1+0.05+10\times0.0001=1.051,$ which gives $2967.53$.
A: $$\frac{1.01^5}{1.01^5-1}=1+\frac{1}{1.01^5-1}=1+\frac{1}{(1+\epsilon)^5-1}$$
$$(1+\epsilon)^5-1=\epsilon ^5+5 \epsilon ^4+10 \epsilon ^3+10 \epsilon ^2+5 \epsilon$$ Long division
$$\frac{1}{(1+\epsilon)^5-1}=\frac{1}{5 \epsilon }-\frac{2}{5}+\frac{2 \epsilon }{5}-\frac{\epsilon
   ^2}{5}-\frac{\epsilon ^3}{25}+O\left(\epsilon ^4\right)$$
$$\frac{(1+\epsilon)^5}{(1+\epsilon)^5-1}=\frac{1}{5 \epsilon }+\frac{3}{5}+\frac{2 \epsilon }{5}-\frac{\epsilon
   ^2}{5}-\frac{\epsilon ^3}{25}+O\left(\epsilon ^4\right)$$
$$\frac{1.01^5}{1.01^5-1}\sim\frac {100}5+\frac 3 5+\frac 2 {500}=\frac{5151}{250}=\frac{20604}{1000}=\color{red}{20.604}$$ instead of $20.6039799616\cdots$
