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.
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.