Answer :
Sure! Let's tackle each part step-by-step:
### Part (a)
Expression: [tex]\((2 - x)^{11}\)[/tex]
Term of interest: The term in [tex]\(x^8\)[/tex]
We use the Binomial Theorem, which states:
[tex]\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \][/tex]
Here, [tex]\(a = 2\)[/tex] and [tex]\(b = -x\)[/tex], [tex]\(n = 11\)[/tex], and we are interested in the term where the power of [tex]\(x\)[/tex] is 8, i.e., [tex]\(x^8\)[/tex].
So, [tex]\(k = 8\)[/tex]. The coefficient is given by:
[tex]\[ \binom{11}{8} \cdot (2)^{11-8} \cdot (-x)^8 \][/tex]
First, calculate [tex]\(\binom{11}{8}\)[/tex]:
[tex]\[ \binom{11}{8} = \binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165 \][/tex]
Then calculate the coefficient:
[tex]\[ 165 \cdot (2)^3 \cdot (-x)^8 = 165 \cdot 8 \cdot x^8 = 1320 x^8 \][/tex]
So, the coefficient of the term in [tex]\(x^8\)[/tex] is 1320.
### Part (b)
Expression: [tex]\((3x + 2)^{14}\)[/tex]
Term of interest: The 13th term in descending powers of [tex]\(x\)[/tex]
In the binomial expansion, the [tex]\(k\)[/tex]-th term (1-indexed) in descending powers of [tex]\(x\)[/tex] corresponds to the [tex]\(n-k+1\)[/tex]-th term in ascending powers of [tex]\(x\)[/tex].
For [tex]\(n = 14\)[/tex] and the 13th term in descending powers, we have:
[tex]\[ k = 14 - (13 - 1) = 2 \][/tex]
The term of interest is:
[tex]\[ \binom{14}{2} (3x)^{14-2} (2)^2 = \binom{14}{2} (3x)^{12} \cdot 4 \][/tex]
First, calculate [tex]\(\binom{14}{2}\)[/tex]:
[tex]\[ \binom{14}{2} = \frac{14 \times 13}{2 \times 1} = 91 \][/tex]
Then calculate the coefficient:
[tex]\[ 91 \cdot (3^{12} \cdot 4 ) \][/tex]
So, we need [tex]\(3^{12}\)[/tex] :
[tex]\[ 3^{12} = 531441 \][/tex]
Thus:
[tex]\[ 91 \cdot 531441 \cdot 4 = 19399344 \][/tex]
So, the coefficient of the 13th term in descending powers of [tex]\(x\)[/tex] is 19399344.
### Part (c)
Expression: [tex]\(\left(2x^3 - \frac{1}{x} \right)^{16}\)[/tex]
Term of interest: Constant term
We need to find the coefficient of the constant term. The constant term occurs where:
[tex]\[ 3k - (16 - k) = 0 \implies 4k - 16 = 0 \implies k = 4 \][/tex]
So the coefficient is:
[tex]\[ \binom{16}{4} (2x^3)^4 \left( -\frac{1}{x} \right)^{12} \][/tex]
First, calculate [tex]\(\binom{16}{4}\)[/tex]:
[tex]\[ \binom{16}{4} = \frac{16 \times 15 \times 14 \times 13}{4 \times 3 \times 2 \times 1} = 1820 \][/tex]
The term is:
[tex]\[ 1820 \cdot (2^4 x^{12}) \cdot (-1)^{12} \cdot x^{-12} \][/tex]
Simplify:
[tex]\[ 1820 \cdot 16 \cdot x^0 = 1820 \cdot 16 = 29120 \][/tex]
Therefore, the constant term's coefficient is 29120.
### Part (d)
Expression: [tex]\((3a + \sqrt{b})^4\)[/tex]
Term of interest: The term in [tex]\(a^2 b\)[/tex]
We use the Binomial Theorem:
[tex]\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \][/tex]
Here, [tex]\(a = 3a\)[/tex], [tex]\(b = \sqrt{b}\)[/tex], [tex]\(n = 4\)[/tex], and we want the term in [tex]\(a^2 b\)[/tex]. The term in [tex]\(a^2 b\)[/tex] corresponds to:
[tex]\[ (3a)^2 (\sqrt{b})^2 \][/tex]
So, we need [tex]\(k=2\)[/tex]. Then the coefficient is:
[tex]\[ \binom{4}{2} (3a)^{4-2} (\sqrt{b})^2 \][/tex]
Calculate [tex]\(\binom{4}{2}\)[/tex]:
[tex]\[ \binom{4}{2} = 6 \][/tex]
Then:
[tex]\[ 6 \cdot (3a)^2 \cdot b = 6 \cdot 9a^2 \cdot b = 54 a^2 b \][/tex]
So, the coefficient of the term in [tex]\(a^2 b\)[/tex] is 54.
### Part (a)
Expression: [tex]\((2 - x)^{11}\)[/tex]
Term of interest: The term in [tex]\(x^8\)[/tex]
We use the Binomial Theorem, which states:
[tex]\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \][/tex]
Here, [tex]\(a = 2\)[/tex] and [tex]\(b = -x\)[/tex], [tex]\(n = 11\)[/tex], and we are interested in the term where the power of [tex]\(x\)[/tex] is 8, i.e., [tex]\(x^8\)[/tex].
So, [tex]\(k = 8\)[/tex]. The coefficient is given by:
[tex]\[ \binom{11}{8} \cdot (2)^{11-8} \cdot (-x)^8 \][/tex]
First, calculate [tex]\(\binom{11}{8}\)[/tex]:
[tex]\[ \binom{11}{8} = \binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165 \][/tex]
Then calculate the coefficient:
[tex]\[ 165 \cdot (2)^3 \cdot (-x)^8 = 165 \cdot 8 \cdot x^8 = 1320 x^8 \][/tex]
So, the coefficient of the term in [tex]\(x^8\)[/tex] is 1320.
### Part (b)
Expression: [tex]\((3x + 2)^{14}\)[/tex]
Term of interest: The 13th term in descending powers of [tex]\(x\)[/tex]
In the binomial expansion, the [tex]\(k\)[/tex]-th term (1-indexed) in descending powers of [tex]\(x\)[/tex] corresponds to the [tex]\(n-k+1\)[/tex]-th term in ascending powers of [tex]\(x\)[/tex].
For [tex]\(n = 14\)[/tex] and the 13th term in descending powers, we have:
[tex]\[ k = 14 - (13 - 1) = 2 \][/tex]
The term of interest is:
[tex]\[ \binom{14}{2} (3x)^{14-2} (2)^2 = \binom{14}{2} (3x)^{12} \cdot 4 \][/tex]
First, calculate [tex]\(\binom{14}{2}\)[/tex]:
[tex]\[ \binom{14}{2} = \frac{14 \times 13}{2 \times 1} = 91 \][/tex]
Then calculate the coefficient:
[tex]\[ 91 \cdot (3^{12} \cdot 4 ) \][/tex]
So, we need [tex]\(3^{12}\)[/tex] :
[tex]\[ 3^{12} = 531441 \][/tex]
Thus:
[tex]\[ 91 \cdot 531441 \cdot 4 = 19399344 \][/tex]
So, the coefficient of the 13th term in descending powers of [tex]\(x\)[/tex] is 19399344.
### Part (c)
Expression: [tex]\(\left(2x^3 - \frac{1}{x} \right)^{16}\)[/tex]
Term of interest: Constant term
We need to find the coefficient of the constant term. The constant term occurs where:
[tex]\[ 3k - (16 - k) = 0 \implies 4k - 16 = 0 \implies k = 4 \][/tex]
So the coefficient is:
[tex]\[ \binom{16}{4} (2x^3)^4 \left( -\frac{1}{x} \right)^{12} \][/tex]
First, calculate [tex]\(\binom{16}{4}\)[/tex]:
[tex]\[ \binom{16}{4} = \frac{16 \times 15 \times 14 \times 13}{4 \times 3 \times 2 \times 1} = 1820 \][/tex]
The term is:
[tex]\[ 1820 \cdot (2^4 x^{12}) \cdot (-1)^{12} \cdot x^{-12} \][/tex]
Simplify:
[tex]\[ 1820 \cdot 16 \cdot x^0 = 1820 \cdot 16 = 29120 \][/tex]
Therefore, the constant term's coefficient is 29120.
### Part (d)
Expression: [tex]\((3a + \sqrt{b})^4\)[/tex]
Term of interest: The term in [tex]\(a^2 b\)[/tex]
We use the Binomial Theorem:
[tex]\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \][/tex]
Here, [tex]\(a = 3a\)[/tex], [tex]\(b = \sqrt{b}\)[/tex], [tex]\(n = 4\)[/tex], and we want the term in [tex]\(a^2 b\)[/tex]. The term in [tex]\(a^2 b\)[/tex] corresponds to:
[tex]\[ (3a)^2 (\sqrt{b})^2 \][/tex]
So, we need [tex]\(k=2\)[/tex]. Then the coefficient is:
[tex]\[ \binom{4}{2} (3a)^{4-2} (\sqrt{b})^2 \][/tex]
Calculate [tex]\(\binom{4}{2}\)[/tex]:
[tex]\[ \binom{4}{2} = 6 \][/tex]
Then:
[tex]\[ 6 \cdot (3a)^2 \cdot b = 6 \cdot 9a^2 \cdot b = 54 a^2 b \][/tex]
So, the coefficient of the term in [tex]\(a^2 b\)[/tex] is 54.
