Answer :
Let's analyze and solve the given problems step-by-step using the provided mathematical approach:
### 1(a) The Binomial Expansion
We need to find the first three terms of the binomial expansion of [tex]\(\left(2 - \frac{x}{8}\right)^{10}\)[/tex] in ascending powers of [tex]\(x\)[/tex].
The general binomial expansion for [tex]\((a + b)^n\)[/tex] is given by:
[tex]\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \][/tex]
In this case, [tex]\(a = 2\)[/tex], [tex]\(b = -\frac{x}{8}\)[/tex], and [tex]\(n = 10\)[/tex]. Let's compute the first three terms:
#### First Term (k = 0):
[tex]\[ \binom{10}{0} (2)^{10-0} \left(-\frac{x}{8}\right)^0 = \binom{10}{0} 2^{10} (1) = 1 \times 1024 \times 1 = 1024 \][/tex]
#### Second Term (k = 1):
[tex]\[ \binom{10}{1} (2)^{10-1} \left(-\frac{x}{8}\right)^1 = 10 \times 2^9 \times \left(-\frac{x}{8}\right) = 10 \times 512 \times \left(-\frac{x}{8}\right) = 10 \times 512 \times -\frac{x}{8} = -640x \][/tex]
#### Third Term (k = 2):
[tex]\[ \binom{10}{2} (2)^{10-2} \left(-\frac{x}{8}\right)^2 = 45 \times 2^8 \times \left(\frac{x^2}{64}\right) = 45 \times 256 \times \frac{x^2}{64} = 45 \times 4 \times x^2 = 180x^2 \][/tex]
So, the first three terms in the binomial expansion of [tex]\(\left(2 - \frac{x}{8}\right)^{10}\)[/tex] in ascending powers of [tex]\(x\)[/tex] are:
[tex]\[ 1024 - 640x + 180x^2 \][/tex]
### 1(b) Finding [tex]\(a\)[/tex]
Given that [tex]\(a\)[/tex] is the constant term in the series expansion of [tex]\(f(x) = \left(2 - \frac{x}{8}\right)^{10}(a + bx)\)[/tex], we can directly use the information provided:
Given: The first term in the series expansion of [tex]\(f(x)\)[/tex] is 256.
The term [tex]\(a\)[/tex] in the expansion corresponds to the constant term of the expression [tex]\(\left(2 - \frac{x}{8}\right)^{10}\)[/tex]. Given that the first term (the constant term) is 1024 in the expansion and given in the problem that the first term of [tex]\(f(x)\)[/tex] is 256:
[tex]\[ 1024a = 256 \][/tex]
Solving for [tex]\(a\)[/tex]:
[tex]\[ a = \frac{256}{1024} = \frac{1}{4} \][/tex]
### 1(c) Finding [tex]\(b\)[/tex]
Given that the second term is [tex]\(352x\)[/tex], let's relate it to the binomial expansion we found earlier. This term corresponds to the coefficient of [tex]\(x\)[/tex] in the product [tex]\(\left(2 - \frac{x}{8}\right)^{10}(a + bx)\)[/tex]:
The coefficient of [tex]\(x\)[/tex] in the expansion is given by:
[tex]\[ -640 + 1024b \][/tex]
Given that this term equals [tex]\(352x\)[/tex],
[tex]\[ -640 + 1024b = 352 \][/tex]
Solving for [tex]\(b\)[/tex]:
[tex]\[ 1024b = 352 + 640 = 992 \][/tex]
[tex]\[ b = \frac{992}{1024} = \frac{31}{32} \][/tex]
Therefore, the values are:
[tex]\[ a = 256, \quad b = 352 \][/tex]
### 1(a) The Binomial Expansion
We need to find the first three terms of the binomial expansion of [tex]\(\left(2 - \frac{x}{8}\right)^{10}\)[/tex] in ascending powers of [tex]\(x\)[/tex].
The general binomial expansion for [tex]\((a + b)^n\)[/tex] is given by:
[tex]\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \][/tex]
In this case, [tex]\(a = 2\)[/tex], [tex]\(b = -\frac{x}{8}\)[/tex], and [tex]\(n = 10\)[/tex]. Let's compute the first three terms:
#### First Term (k = 0):
[tex]\[ \binom{10}{0} (2)^{10-0} \left(-\frac{x}{8}\right)^0 = \binom{10}{0} 2^{10} (1) = 1 \times 1024 \times 1 = 1024 \][/tex]
#### Second Term (k = 1):
[tex]\[ \binom{10}{1} (2)^{10-1} \left(-\frac{x}{8}\right)^1 = 10 \times 2^9 \times \left(-\frac{x}{8}\right) = 10 \times 512 \times \left(-\frac{x}{8}\right) = 10 \times 512 \times -\frac{x}{8} = -640x \][/tex]
#### Third Term (k = 2):
[tex]\[ \binom{10}{2} (2)^{10-2} \left(-\frac{x}{8}\right)^2 = 45 \times 2^8 \times \left(\frac{x^2}{64}\right) = 45 \times 256 \times \frac{x^2}{64} = 45 \times 4 \times x^2 = 180x^2 \][/tex]
So, the first three terms in the binomial expansion of [tex]\(\left(2 - \frac{x}{8}\right)^{10}\)[/tex] in ascending powers of [tex]\(x\)[/tex] are:
[tex]\[ 1024 - 640x + 180x^2 \][/tex]
### 1(b) Finding [tex]\(a\)[/tex]
Given that [tex]\(a\)[/tex] is the constant term in the series expansion of [tex]\(f(x) = \left(2 - \frac{x}{8}\right)^{10}(a + bx)\)[/tex], we can directly use the information provided:
Given: The first term in the series expansion of [tex]\(f(x)\)[/tex] is 256.
The term [tex]\(a\)[/tex] in the expansion corresponds to the constant term of the expression [tex]\(\left(2 - \frac{x}{8}\right)^{10}\)[/tex]. Given that the first term (the constant term) is 1024 in the expansion and given in the problem that the first term of [tex]\(f(x)\)[/tex] is 256:
[tex]\[ 1024a = 256 \][/tex]
Solving for [tex]\(a\)[/tex]:
[tex]\[ a = \frac{256}{1024} = \frac{1}{4} \][/tex]
### 1(c) Finding [tex]\(b\)[/tex]
Given that the second term is [tex]\(352x\)[/tex], let's relate it to the binomial expansion we found earlier. This term corresponds to the coefficient of [tex]\(x\)[/tex] in the product [tex]\(\left(2 - \frac{x}{8}\right)^{10}(a + bx)\)[/tex]:
The coefficient of [tex]\(x\)[/tex] in the expansion is given by:
[tex]\[ -640 + 1024b \][/tex]
Given that this term equals [tex]\(352x\)[/tex],
[tex]\[ -640 + 1024b = 352 \][/tex]
Solving for [tex]\(b\)[/tex]:
[tex]\[ 1024b = 352 + 640 = 992 \][/tex]
[tex]\[ b = \frac{992}{1024} = \frac{31}{32} \][/tex]
Therefore, the values are:
[tex]\[ a = 256, \quad b = 352 \][/tex]
