Thread: The coefficient of x^10 in this expansion

1. The coefficient of x^10 in this expansion

Hi, I've been doing Binomial Theorem, and here is an annoying question I encountered:

Find the coefficient of $\displaystyle x^{10}$ in the expansion of:
$\displaystyle \left ( x - \frac{1}{x} \right )^6 \left ( x+\frac{1}{x} \right )^8$

Thanks.

2. Originally Posted by differentiate
Hi, I've been doing Binomial Theorem, and here is an annoying question I encountered:

Find the coefficient of $\displaystyle x^{10}$ in the expansion of:
$\displaystyle \left ( x - \frac{1}{x} \right )^6 \left ( x+\frac{1}{x} \right )^8$

Thanks.
The general term is $\displaystyle {6 \choose r} x^r \left(- \frac{1}{x} \right)^{6 - r} \cdot {8 \choose t} x^t \left(\frac{1}{x} \right)^{8 - t} = (-1)^{6 - r} {6 \choose r} {8 \choose t} x^{r + r - 6} x^{t + t - 8}$

$\displaystyle = (-1)^{6 - r} {6 \choose r} {8 \choose t} x^{2r + 2t - 14}$.

You require the power of x to be 10. Therefore:

$\displaystyle 2r + 2t - 14 = 10 \Rightarrow r + t = 12$.

Construct the required coefficient from the values of r and t that satisfy the above equation.

3. OMG you're a genius!! THANK YOU!!!!!!!!!!!!!!!!!!!!! Your method works much better than the one described in the textbook!!!!!!!!!!!!!!!!!!!!!!! YOU GENIUS!!

4. Originally Posted by differentiate
OMG you're a genius!! THANK YOU!!!!!!!!!!!!!!!!!!!!! Your method works much better than the one described in the textbook!!!!!!!!!!!!!!!!!!!!!!! YOU GENIUS!!
No doubt someone will post a better answer that will make me look somewhat less than a genius. (I was going to first simplify the expression using the difference of two squares formula but someone else might like to expand (ha ha) upon that).

5. Originally Posted by mr fantastic
No doubt someone will post a better answer that will make me look somewhat less than a genius.
No one would dare!

(I was going to first simplify the expression using the difference of two squares formula but someone else might like to expand (ha ha) upon that).