To find the 4th term of the expansion of [tex]\((1 - 2x)^n\)[/tex], we need to use the binomial theorem. The binomial theorem states that:
[tex]\[
(a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k
\][/tex]
For the given expression [tex]\((1 - 2x)^n\)[/tex], we can identify [tex]\(a\)[/tex] and [tex]\(b\)[/tex] as follows:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = -2x\)[/tex]
The 4th term in the binomial expansion corresponds to [tex]\(k = 3\)[/tex] (since we start counting from [tex]\(k = 0\)[/tex]).
The general term in the expansion is given by:
[tex]\[
\binom{n}{k} a^{n-k} b^k
\][/tex]
Substituting [tex]\(a = 1\)[/tex], [tex]\(b = -2x\)[/tex], and [tex]\(k = 3\)[/tex] into the formula, we get:
[tex]\[
\text{4th term} = \binom{n}{3} \cdot (1)^{n-3} \cdot (-2x)^3
\][/tex]
Since [tex]\((1)^{n-3}\)[/tex] is always 1, the expression simplifies to:
[tex]\[
\text{4th term} = \binom{n}{3} \cdot (-2x)^3
\][/tex]
Now we need the binomial coefficient [tex]\(\binom{n}{3}\)[/tex]. According to the provided row of Pascal's triangle [tex]\([1, 6, 15, 20, 15, 6, 1]\)[/tex], the fourth value (corresponding to [tex]\(k = 3\)[/tex]) is [tex]\(20\)[/tex].
So substituting [tex]\(\binom{n}{3} = 20\)[/tex], we get:
[tex]\[
\text{4th term} = 20 \cdot (-2x)^3
\][/tex]
Next, simplify the term [tex]\((-2x)^3\)[/tex]:
[tex]\[
(-2x)^3 = (-2)^3 \cdot x^3 = -8 \cdot x^3
\][/tex]
Combining these results:
[tex]\[
\text{4th term} = 20 \cdot (-8x^3) = 20 \cdot -8 \cdot x^3 = -160x^3
\][/tex]
Thus, the 4th term of the expansion of [tex]\((1 - 2x)^n\)[/tex] is:
[tex]\[
-160x^3
\][/tex]
So the correct answer is [tex]\(\boxed{-160x^3}\)[/tex].
