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.

Credits: MF15
Credits: MF15

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}



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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