How do you find the coefficient of a of the term az^4t^8 in the expansion of the binomial (z^2-t)^10?

Aug 3, 2017

$a = 45.$

Explanation:

We know that, the General ${\left(r + 1\right)}^{t h}$ Term, denoted by,

${T}_{r + 1} ,$ 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.

In our Problem, $a = {z}^{2} , b = - t , n = 10.$

:. T_(r+1)=""_10C_r(z^2)^(10-r)(-t)^r,

=""_10C_r*z^(20-2r)(-1)^r*t^r, r=0,1,2,...,10.

Comparing this with, $a {z}^{4} {t}^{8} ,$ we have,

a=(-1)^r""_10C_r, 20-2r=4, and, r=8.

$\therefore a = {\left(- 1\right)}^{8} {\text{_10C_8=}}_{10} {C}_{2} = \frac{10 \cdot 9}{1 \cdot 2} = 45.$