To expand and simplify the binomial expression [tex]\(\left(1 - \frac{x}{2}\right)^5\)[/tex], we can 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 expression [tex]\(\left(1 - \frac{x}{2}\right)^5\)[/tex], we identify [tex]\(a = 1\)[/tex], [tex]\(b = -\frac{x}{2}\)[/tex], and [tex]\(n = 5\)[/tex]. Applying the binomial theorem, we get:
[tex]\[
\left(1 - \frac{x}{2}\right)^5 = \sum_{k=0}^{5} \binom{5}{k} \left(1\right)^{5-k} \left(-\frac{x}{2}\right)^k
\][/tex]
We can break this down into individual terms. Calculating each term:
- For [tex]\(k = 0\)[/tex]:
[tex]\[
\binom{5}{0} \left(1\right)^5 \left(-\frac{x}{2}\right)^0 = 1 \cdot 1 \cdot 1 = 1
\][/tex]
- For [tex]\(k = 1\)[/tex]:
[tex]\[
\binom{5}{1} \left(1\right)^4 \left(-\frac{x}{2}\right)^1 = 5 \cdot 1 \cdot \left(-\frac{x}{2}\right) = -\frac{5x}{2}
\][/tex]
- For [tex]\(k = 2\)[/tex]:
[tex]\[
\binom{5}{2} \left(1\right)^3 \left(-\frac{x}{2}\right)^2 = 10 \cdot 1 \cdot \frac{x^2}{4} = \frac{10x^2}{4} = \frac{5x^2}{2}
\][/tex]
- For [tex]\(k = 3\)[/tex]:
[tex]\[
\binom{5}{3} \left(1\right)^2 \left(-\frac{x}{2}\right)^3 = 10 \cdot 1 \cdot \left(-\frac{x}{2}\right)^3 = 10 \cdot \left(-\frac{x^3}{8}\right) = -\frac{10x^3}{8} = -\frac{5x^3}{4}
\][/tex]
- For [tex]\(k = 4\)[/tex]:
[tex]\[
\binom{5}{4} \left(1\right)^1 \left(-\frac{x}{2}\right)^4 = 5 \cdot 1 \cdot \frac{x^4}{16} = \frac{5x^4}{16}
\][/tex]
- For [tex]\(k = 5\)[/tex]:
[tex]\[
\binom{5}{5} \left(1\right)^0 \left(-\frac{x}{2}\right)^5 = 1 \cdot 1 \cdot \left(-\frac{x}{2}\right)^5 = -\frac{x^5}{32}
\][/tex]
Now, summing up all these terms, we get the expanded binomial expression:
[tex]\[
\left(1 - \frac{x}{2}\right)^5 = 1 - \frac{5x}{2} + \frac{5x^2}{2} - \frac{5x^3}{4} + \frac{5x^4}{16} - \frac{x^5}{32}
\][/tex]
So the expanded and simplified form of the binomial expression [tex]\(\left(1 - \frac{x}{2}\right)^5\)[/tex] is:
[tex]\[
-x^5/32 + 5x^4/16 - 5x^3/4 + 5x^2/2 - 5x/2 + 1
\][/tex]
