# How can I find the coefficient of x^12

1. Nov 23, 2004

### DeathKnight

How can I find the coefficient of x^12 in:
$$(2+x)^{14}(1+ \frac{2}{x})^{14}$$
I did it like that:
$$(2+x)^{14}(1+ \frac{2}{x})^{14}$$
Can be written as:
$$(4+ \frac{4+x^2}{x})^{14}$$
Now the term $$T_{r+1}$$ can be represented as:
$$14C_r (4)^{14-r}(\frac{4+x^2}{x})^r$$ (How can I write 14Cr?)
Dont know where to go from here?

Thanks in advance for any help.

Last edited: Nov 23, 2004
2. Nov 23, 2004

### matt grime

There's more than one way to skin a cat:

how can you obtain x^12 from the two original brackets? You need x^14 from one and x^{-2} from the other and... add 'em up.

for the tex for binomial coeffs click on:

$$\binom{n}{r}$$

3. Nov 23, 2004

### DeathKnight

Thanks alot for your reply Matt. I got what you said but I'm still curious to find out how can I find the answer to it by using the method I stated above? I have done many similar questions and I know I have to find the value of r for which the power of x is 12. Is this possible?

Last edited: Nov 23, 2004
4. Nov 23, 2004

### matt grime

of course it is- just not very easy, you just need, for each r to find the positive integer s where 0<=s<=r, if it exists, such that 2s-r=12 and then add up, for each one where this happens,

$$\binom{14}{r}\binom{r}{s}4^{14-r}4^{r-s}$$

hardly elegant but then you made the problem much harder with your choice of "simplification"