Alg10.6

Find the independent term of $x\,$ in the expansion of $(x-\frac{1}{x})^6\,$

$T_{r+1}=6\!C_r\cdot (x)^{6-r}\cdot (\frac{-1}{x})^r\,$

$6\!C_r\cdot x^{6-r}\cdot \frac{(-1)^r}{x^r}\,$

$6\!C_r \cdot x^{6-2r} \cdot(-1)^r\,$

To get the value independent of $x\,$ we have to find r so that $6-2r=0,r=3\,$

Therefore,the 4th term is independent of x

Hence $T_{3+1}=9\!C_3\cdot (-1)^3=-9\!C_3\,$