Sure, let's solve each part step-by-step.
### Part 4.1: Finding the Middle Term in the Expansion of [tex]\((3x - y)^4\)[/tex]
We use the binomial theorem for the expansion:
[tex]\[
(a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k
\][/tex]
Here, [tex]\(a = 3x\)[/tex], [tex]\(b = -y\)[/tex], and [tex]\(n = 4\)[/tex].
To find the middle term in the expansion of [tex]\((3x - y)^4\)[/tex]:
- Since [tex]\( n = 4 \)[/tex] is even, the middle term is the [tex]\((\frac{n}{2} + 1)\)[/tex]-th term or the [tex]\((\frac{4}{2} + 1 = 3)\)[/tex]-rd term in the expansion.
The [tex]\((k+1)\)[/tex]-th term of the expansion is given by:
[tex]\[
T_{k+1} = \binom{n}{k} (3x)^{n-k} (-y)^k
\][/tex]
For the middle term, [tex]\(k = \frac{4}{2} = 2\)[/tex]:
[tex]\[
T_{3} = \binom{4}{2} (3x)^{4-2} (-y)^2
\][/tex]
Calculating each component:
- [tex]\(\binom{4}{2} = \frac{4!}{2!2!} = 6\)[/tex]
- [tex]\((3x)^2 = 9x^2\)[/tex]
- [tex]\((-y)^2 = y^2\)[/tex]
Combining these together:
[tex]\[
T_{3} = 6 \cdot 9x^2 \cdot y^2 = 54x^2 y^2
\][/tex]
So, the middle term is:
[tex]\[
54x^2 y^2
\][/tex]
### Part 4.2: Finding the Term that Contains [tex]\(x^5\)[/tex] in the Expansion of [tex]\((2x + 3y)^8\)[/tex]
We use the binomial theorem for the expansion:
[tex]\[
(a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k
\][/tex]
Here, [tex]\(a = 2x\)[/tex], [tex]\(b = 3y\)[/tex], and [tex]\(n = 8\)[/tex].
We are asked for the term that contains [tex]\(x^5\)[/tex]. In the binomial expansion, the [tex]\((k+1)\)[/tex]-th term is:
[tex]\[
T_{k+1} = \binom{n}{k} (2x)^{n-k} (3y)^k
\][/tex]
We need to find [tex]\(k\)[/tex] so that the power of [tex]\(x\)[/tex] is 5:
[tex]\[
(2x)^{8-k} \implies (8-k) = 5 \implies k = 3
\][/tex]
The term when [tex]\(k = 3\)[/tex] is:
[tex]\[
T_{4} = \binom{8}{3} (2x)^{8-3} (3y)^3 = \binom{8}{3} (2x)^5 (3y)^3
\][/tex]
Calculating each component:
- [tex]\(\binom{8}{3} = \frac{8!}{3!5!} = 56\)[/tex]
- [tex]\((2x)^5 = 32x^5\)[/tex]
- [tex]\((3y)^3 = 27y^3\)[/tex]
Combining these together:
[tex]\[
T_{4} = 56 \cdot 32x^5 \cdot 27y^3
\][/tex]
[tex]\[
T_{4} = 56 \cdot 864 x^5 y^3 = 48384 x^5 y^3
\][/tex]
So, the term that contains [tex]\(x^5\)[/tex] is:
[tex]\[
48384 x^5 y^3
\][/tex]
### Summary
- 4.1: The middle term in the expansion of [tex]\((3x - y)^4\)[/tex] is [tex]\(54x^2 y^2\)[/tex].
- 4.2: The term that contains [tex]\(x^5\)[/tex] in the expansion of [tex]\((2x + 3y)^8\)[/tex] is [tex]\(48384 x^5 y^3\)[/tex].
