ANSWER
[tex]T_{5} =4480 {x}^{3} [/tex]
EXPLANATION
We want to find the fifth term of the binomial expansion
[tex](2x - 2)^{7} [/tex]
When we compare to
[tex] {(a + b)}^{n} [/tex]
we have n=7, a=2x and b=-2
For the 5th term,
[tex]r + 1 = 5[/tex]
This means that,
[tex]r = 4[/tex]
The fifth term can be found using the formula,
[tex]T_{r+1} = ^nC_r {a}^{n - r} {b}^{r} [/tex]
[tex]T_{5} = ^7C_4 {(2x)}^{7 - 4} {( - 2)}^{4} [/tex]
We substitute the values to obtain,
[tex]T_{5} = ^7C_4 {(2x)}^{3} {( - 2)}^{4} [/tex]
[tex]T_{5} = 35 \times {8x}^{3} \times 16[/tex]
[tex]T_{5} =4480 {x}^{3} [/tex]
