Answer :
To solve the problem, let's start by clearly outlining both parts of the problem and their step-by-step solutions.
### Part (a): Expanding [tex]\((2 - 3x)^4\)[/tex]
First, we need to expand [tex]\((2 - 3x)^4\)[/tex]. We can use the binomial theorem for this expansion. The binomial theorem states:
[tex]\[ (a - b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} (-b)^k \][/tex]
Here, [tex]\(a = 2\)[/tex], [tex]\(b = 3x\)[/tex], and [tex]\(n = 4\)[/tex].
[tex]\[ (2 - 3x)^4 = \sum_{k=0}^{4} \binom{4}{k} (2)^{4-k} (-3x)^k \][/tex]
Let's compute each term one by one:
1. [tex]\(k = 0\)[/tex]
[tex]\[ \binom{4}{0} (2)^{4-0} (-3x)^0 = 1 \cdot 2^4 \cdot (3x)^0 = 1 \cdot 16 \cdot 1 = 16 \][/tex]
2. [tex]\(k = 1\)[/tex]
[tex]\[ \binom{4}{1} (2)^{4-1} (-3x)^1 = 4 \cdot 2^3 \cdot (-3x) = 4 \cdot 8 \cdot (-3x) = 4 \cdot 8 \cdot (-3x) = -96x \][/tex]
3. [tex]\(k = 2\)[/tex]
[tex]\[ \binom{4}{2} (2)^{4-2} (-3x)^2 = 6 \cdot 2^2 \cdot (9x^2) = 6 \cdot 4 \cdot 9x^2 = 216x^2 \][/tex]
4. [tex]\(k = 3\)[/tex]
[tex]\[ \binom{4}{3} (2)^{4-3} (-3x)^3 = 4 \cdot 2 \cdot (-27x^3) = 4 \cdot 2 \cdot (-27x^3) = -216x^3 \][/tex]
5. [tex]\(k = 4\)[/tex]
[tex]\[ \binom{4}{4} (2)^{4-4} (-3x)^4 = 1 \cdot 1 \cdot 81x^4 = 81x^4 \][/tex]
Combining these terms, we get:
[tex]\[ (2 - 3x)^4 = 16 - 96x + 216x^2 - 216x^3 + 81x^4 \][/tex]
Thus, the coefficients of the expansion are [tex]\(16\)[/tex], [tex]\(-96\)[/tex], [tex]\(216\)[/tex], [tex]\(-216\)[/tex], and [tex]\(81\)[/tex].
### Part (b): Finding [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]
Next, we need to determine the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] given that the sum of the first three terms in ascending powers of [tex]\(x\)[/tex] in the expansion of [tex]\((2 - 3x)^4\left(1 + \frac{a}{x}\right)\)[/tex] is [tex]\(\frac{32}{x} + b + cx\)[/tex].
First, consider the product [tex]\((2 - 3x)^4 \cdot \left(1 + \frac{a}{x}\right)\)[/tex]:
[tex]\[ [(2 - 3x)^4] \cdot \left(1 + \frac{a}{x}\right) \][/tex]
We know the expansion of [tex]\((2 - 3x)^4\)[/tex] and we will multiply each term by [tex]\(\left(1 + \frac{a}{x}\right)\)[/tex]. We focus on the sum of the first three terms in ascending powers of [tex]\(x\)[/tex].
The term for [tex]\(x^{-1}\)[/tex] comes from:
1. [tex]\(16 \times \frac{a}{x} = \frac{16a}{x}\)[/tex]
Matching this with [tex]\(\frac{32}{x}\)[/tex], we get:
[tex]\[ \frac{16a}{x} = \frac{32}{x} \implies 16a = 32 \implies a = 2 \][/tex]
For the constant term:
1. From [tex]\((2 - 3x)^4\)[/tex], the term is [tex]\(16\)[/tex].
2. No contribution from the [tex]\( \frac{a}{x} \)[/tex] portion because there is no [tex]\( \frac{1}{x} \)[/tex] in the constant term.
Therefore, [tex]\(b = 16\)[/tex].
For the [tex]\(x\)[/tex] coefficient term:
1. From [tex]\((2 - 3x)^4\)[/tex], the term is [tex]\(-96x\)[/tex].
2. From [tex]\(16 \times \frac{2}{x}\)[/tex], this is not contributing because it would result in a constant.
Combining these, we get [tex]\( b = 16\)[/tex].
From the expansion and comparison, we got:
[tex]\[ c = -96 \][/tex]
Thus, the values of each of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are:
[tex]\[ a = 2, \quad b = 16, \quad c = -96 \][/tex]
These were found to closely match the results calculated.
### Part (a): Expanding [tex]\((2 - 3x)^4\)[/tex]
First, we need to expand [tex]\((2 - 3x)^4\)[/tex]. We can use the binomial theorem for this expansion. The binomial theorem states:
[tex]\[ (a - b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} (-b)^k \][/tex]
Here, [tex]\(a = 2\)[/tex], [tex]\(b = 3x\)[/tex], and [tex]\(n = 4\)[/tex].
[tex]\[ (2 - 3x)^4 = \sum_{k=0}^{4} \binom{4}{k} (2)^{4-k} (-3x)^k \][/tex]
Let's compute each term one by one:
1. [tex]\(k = 0\)[/tex]
[tex]\[ \binom{4}{0} (2)^{4-0} (-3x)^0 = 1 \cdot 2^4 \cdot (3x)^0 = 1 \cdot 16 \cdot 1 = 16 \][/tex]
2. [tex]\(k = 1\)[/tex]
[tex]\[ \binom{4}{1} (2)^{4-1} (-3x)^1 = 4 \cdot 2^3 \cdot (-3x) = 4 \cdot 8 \cdot (-3x) = 4 \cdot 8 \cdot (-3x) = -96x \][/tex]
3. [tex]\(k = 2\)[/tex]
[tex]\[ \binom{4}{2} (2)^{4-2} (-3x)^2 = 6 \cdot 2^2 \cdot (9x^2) = 6 \cdot 4 \cdot 9x^2 = 216x^2 \][/tex]
4. [tex]\(k = 3\)[/tex]
[tex]\[ \binom{4}{3} (2)^{4-3} (-3x)^3 = 4 \cdot 2 \cdot (-27x^3) = 4 \cdot 2 \cdot (-27x^3) = -216x^3 \][/tex]
5. [tex]\(k = 4\)[/tex]
[tex]\[ \binom{4}{4} (2)^{4-4} (-3x)^4 = 1 \cdot 1 \cdot 81x^4 = 81x^4 \][/tex]
Combining these terms, we get:
[tex]\[ (2 - 3x)^4 = 16 - 96x + 216x^2 - 216x^3 + 81x^4 \][/tex]
Thus, the coefficients of the expansion are [tex]\(16\)[/tex], [tex]\(-96\)[/tex], [tex]\(216\)[/tex], [tex]\(-216\)[/tex], and [tex]\(81\)[/tex].
### Part (b): Finding [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]
Next, we need to determine the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] given that the sum of the first three terms in ascending powers of [tex]\(x\)[/tex] in the expansion of [tex]\((2 - 3x)^4\left(1 + \frac{a}{x}\right)\)[/tex] is [tex]\(\frac{32}{x} + b + cx\)[/tex].
First, consider the product [tex]\((2 - 3x)^4 \cdot \left(1 + \frac{a}{x}\right)\)[/tex]:
[tex]\[ [(2 - 3x)^4] \cdot \left(1 + \frac{a}{x}\right) \][/tex]
We know the expansion of [tex]\((2 - 3x)^4\)[/tex] and we will multiply each term by [tex]\(\left(1 + \frac{a}{x}\right)\)[/tex]. We focus on the sum of the first three terms in ascending powers of [tex]\(x\)[/tex].
The term for [tex]\(x^{-1}\)[/tex] comes from:
1. [tex]\(16 \times \frac{a}{x} = \frac{16a}{x}\)[/tex]
Matching this with [tex]\(\frac{32}{x}\)[/tex], we get:
[tex]\[ \frac{16a}{x} = \frac{32}{x} \implies 16a = 32 \implies a = 2 \][/tex]
For the constant term:
1. From [tex]\((2 - 3x)^4\)[/tex], the term is [tex]\(16\)[/tex].
2. No contribution from the [tex]\( \frac{a}{x} \)[/tex] portion because there is no [tex]\( \frac{1}{x} \)[/tex] in the constant term.
Therefore, [tex]\(b = 16\)[/tex].
For the [tex]\(x\)[/tex] coefficient term:
1. From [tex]\((2 - 3x)^4\)[/tex], the term is [tex]\(-96x\)[/tex].
2. From [tex]\(16 \times \frac{2}{x}\)[/tex], this is not contributing because it would result in a constant.
Combining these, we get [tex]\( b = 16\)[/tex].
From the expansion and comparison, we got:
[tex]\[ c = -96 \][/tex]
Thus, the values of each of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are:
[tex]\[ a = 2, \quad b = 16, \quad c = -96 \][/tex]
These were found to closely match the results calculated.
