To find the first four terms of the binomial expansion of [tex]\(\left(1 - \frac{1}{8} x\right)^{10}\)[/tex] in ascending powers of [tex]\(x\)[/tex], we will use the Binomial Theorem. The Binomial Theorem states:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
In this case, [tex]\( a = 1 \)[/tex], [tex]\( b = -\frac{1}{8} x \)[/tex], and [tex]\( n = 10 \)[/tex]. We will calculate the terms for [tex]\( k = 0, 1, 2, \)[/tex] and [tex]\( 3 \)[/tex].
### Term for [tex]\( k = 0 \)[/tex]:
[tex]\[
\binom{10}{0} (1)^{10-0} \left(-\frac{1}{8} x\right)^0 = \binom{10}{0} (1)^{10} (1) = 1
\][/tex]
### Term for [tex]\( k = 1 \)[/tex]:
[tex]\[
\binom{10}{1} (1)^{10-1} \left(-\frac{1}{8} x\right)^1 = \binom{10}{1} (1)^9 \left(-\frac{1}{8} x\right) = 10 \cdot 1 \cdot \left(-\frac{1}{8} x\right) = -\frac{10}{8} x = -1.25 x
\][/tex]
### Term for [tex]\( k = 2 \)[/tex]:
[tex]\[
\binom{10}{2} (1)^{10-2} \left(-\frac{1}{8} x\right)^2 = \binom{10}{2} (1)^8 \left(-\frac{1}{8} x\right)^2 = \binom{10}{2} \left(\frac{1}{64} x^2\right)
\][/tex]
We need to simplify [tex]\(\binom{10}{2}\)[/tex]:
[tex]\[
\binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45
\][/tex]
Thus, the term is:
[tex]\[
45 \cdot \frac{1}{64} x^2 = \frac{45}{64} x^2 = 0.703125 x^2
\][/tex]
### Term for [tex]\( k = 3 \)[/tex]:
[tex]\[
\binom{10}{3} (1)^{10-3} \left(-\frac{1}{8} x\right)^3 = \binom{10}{3} (1)^7 \left(-\frac{1}{8} x\right)^3 = \binom{10}{3} \left(-\frac{1}{512} x^3\right)
\][/tex]
We need to simplify [tex]\(\binom{10}{3}\)[/tex]:
[tex]\[
\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120
\][/tex]
Thus, the term is:
[tex]\[
120 \cdot -\frac{1}{512} x^3 = -\frac{120}{512} x^3 = -\frac{15}{64} x^3 = -0.234375 x^3
\][/tex]
### Conclusion
The first four terms in the binomial expansion of [tex]\(\left(1 - \frac{1}{8} x\right)^{10}\)[/tex] are:
[tex]\[
1, \quad -1.25 x, \quad 0.703125 x^2, \quad -0.234375 x^3
\][/tex]
