This problem seem to bug many students, especially J1s who are doing Maclaurin’s series now. Many of the students are not sure how to find the coefficient of $x^r$. They wonder how to “see” it. So here, I’ll attempt to show a direct method that doesn’t require us to stare and think.

Consider $(2+x)^{-3}$ and we are interested to find the coefficient of $x^r$.

First of all, to apply $(1+x)^n$ formula, we rewrite to $2^{-3}(1+\frac{x}{2})^{-3}$. This step is based on indices and students should always check that the copied the correct power.

Now consider the formula above that is found in the MF15. We will take the coefficient of $x^r$ that’s found there, taking note that $n = -3$ here.

We will end of with the following

$(2)^{-3}[\frac{(-3)(-4) \ldots (-3-r+1)}{r!}](\frac{x}{2})^{r}$

Take note we must copy the $(2)^{-3}$ alongside too and we substitute the $n$ in and preserve the $r$.

Simplifying, we have

$(2)^{-3}[\frac{(-3)(-4) \ldots (-2-r)}{r!}](\frac{x}{2})^{r}$

$(2)^{-3}[\frac{(-1)^{r}(3)(4) \ldots (r+2)}{r!}](\frac{x}{2})^{r}$

Here we attempt to factorise $(-1)^r$ out, we need to figure how many (-1)’s we have here to factorise. To find out, we can take $(r+2)-3+1 = r$ to find out how many terms there are. Next, we try to simplify the factorials.

$(2)^{-3}[\frac{(-1)^{r}(3)(4) \ldots (r+2)}{r!}](\frac{1}{2})^{r}x^{r}$

$(2)^{-3}[\frac{(-1)^{r}(r+1)(r+2)}{2}](2)^{-r}x^{r}$

$(2)^{-4-r}[(-1)^{r}(r+1)(r+2)]x^{r}$

Therefore, the coefficient is $(2)^{-4-r}[(-1)^{r}(r+1)(r+2)]$.

Note: $(-1)^r$ cannot be further simplified as we do not know if $r$ is even or odd. In the event that $r$ is even, then $(-1)^r = 1$ and if $r$ is odd, then $(-1)^r = -1$. An example is if we consider $(2-x)^{-3}$ instead of $(2+x)^{-3}$, we observe that we will find a $(-1)^{2r}$ which is equals to 1 since $(-1)^2r = ((-1)^2)^r = 1^r = 1$.

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