# How do you find the sixth term of (x-1/2)^10?

Jun 23, 2017

$- \frac{63}{8} {x}^{5.}$

#### Explanation:

We know that, the General ${\left(r + 1\right)}^{t h}$ Term, i.e., ${T}_{r + 1} ,$ in

the expansion of ${\left(a + b\right)}^{n}$ is given by,

T_(r+1)=""_nC_ra^(n-r)b^r, r=0,1,2,...,n.

We have, a=x,b=-1/2, n=10, &, since ${T}_{6}$ is required, $r = 5.$

Accordingly, T_(5+1)=""_10C_5x^(10-5)(-1/2)^5,

$= \frac{\left(10\right) \left(9\right) \left(8\right) \left(7\right) \left(6\right)}{\left(1\right) \left(2\right) \left(3\right) \left(4\right) \left(5\right)} {x}^{5} \left(- \frac{1}{2} ^ 5\right) ,$

$\Rightarrow {T}_{6} = - \frac{63}{8} {x}^{5.}$

Enjoy Maths.!