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Sagot :
Certainly! Let's tackle each part of the question step-by-step:
### Part 4.1: Middle Term in the Expansion of [tex]\((3x - y)^4\)[/tex]
To find the middle term in the expansion of [tex]\((3x - y)^4\)[/tex], we can 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]
For the expression [tex]\((3x - y)^4\)[/tex], we have [tex]\(a = 3x\)[/tex], [tex]\(b = -y\)[/tex], and [tex]\(n = 4\)[/tex].
Since [tex]\(n\)[/tex] is 4, there are 5 terms in the expansion (from [tex]\(k = 0\)[/tex] to [tex]\(k = 4\)[/tex]). The middle term is the one where [tex]\(k\)[/tex] is half of [tex]\(n\)[/tex], which is [tex]\(k = 2\)[/tex].
Use the binomial coefficient and substitute [tex]\(k = 2\)[/tex]:
[tex]\[ \binom{4}{2} (3x)^{4-2} (-y)^2 \][/tex]
Calculate the binomial coefficient:
[tex]\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \][/tex]
Now, substitute the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[ 6 \cdot (3x)^2 \cdot (-y)^2 = 6 \cdot 9x^2 \cdot y^2 = 54x^2y^2 \][/tex]
Thus, the middle term in the expansion of [tex]\((3x - y)^4\)[/tex] is:
[tex]\[ 54x^2y^2 \][/tex]
### Part 4.2: Term with [tex]\(x^5\)[/tex] in the Expansion of [tex]\((2x + 3y)^8\)[/tex]
To find the term that contains [tex]\(x^5\)[/tex] in the expansion of [tex]\((2x + 3y)^8\)[/tex], we also use the binomial theorem. Here, [tex]\(a = 2x\)[/tex], [tex]\(b = 3y\)[/tex], and [tex]\(n = 8\)[/tex].
We need the term where the power of [tex]\(x\)[/tex] is 5. In each term of the expansion, the power of [tex]\(x\)[/tex] is given by [tex]\(n - k\)[/tex], so we set [tex]\(n - k = 5\)[/tex]. Therefore, [tex]\(k = 3\)[/tex].
Use the binomial theorem formula for the [tex]\(k\)[/tex]-th term:
[tex]\[ \binom{8}{3} (2x)^{8-3} (3y)^3 \][/tex]
Calculate the binomial coefficient:
[tex]\[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \][/tex]
Now, substitute the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex], and the powers:
[tex]\[ 56 \cdot (2x)^5 \cdot (3y)^3 = 56 \cdot 32x^5 \cdot 27y^3 = 56 \cdot 864x^5y^3 = 48384x^5y^3 \][/tex]
Thus, the term that contains [tex]\(x^5\)[/tex] in the expansion of [tex]\((2x + 3y)^8\)[/tex] is:
[tex]\[ 48384x^5y^3 \][/tex]
### Part 4.1: Middle Term in the Expansion of [tex]\((3x - y)^4\)[/tex]
To find the middle term in the expansion of [tex]\((3x - y)^4\)[/tex], we can 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]
For the expression [tex]\((3x - y)^4\)[/tex], we have [tex]\(a = 3x\)[/tex], [tex]\(b = -y\)[/tex], and [tex]\(n = 4\)[/tex].
Since [tex]\(n\)[/tex] is 4, there are 5 terms in the expansion (from [tex]\(k = 0\)[/tex] to [tex]\(k = 4\)[/tex]). The middle term is the one where [tex]\(k\)[/tex] is half of [tex]\(n\)[/tex], which is [tex]\(k = 2\)[/tex].
Use the binomial coefficient and substitute [tex]\(k = 2\)[/tex]:
[tex]\[ \binom{4}{2} (3x)^{4-2} (-y)^2 \][/tex]
Calculate the binomial coefficient:
[tex]\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \][/tex]
Now, substitute the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[ 6 \cdot (3x)^2 \cdot (-y)^2 = 6 \cdot 9x^2 \cdot y^2 = 54x^2y^2 \][/tex]
Thus, the middle term in the expansion of [tex]\((3x - y)^4\)[/tex] is:
[tex]\[ 54x^2y^2 \][/tex]
### Part 4.2: Term with [tex]\(x^5\)[/tex] in the Expansion of [tex]\((2x + 3y)^8\)[/tex]
To find the term that contains [tex]\(x^5\)[/tex] in the expansion of [tex]\((2x + 3y)^8\)[/tex], we also use the binomial theorem. Here, [tex]\(a = 2x\)[/tex], [tex]\(b = 3y\)[/tex], and [tex]\(n = 8\)[/tex].
We need the term where the power of [tex]\(x\)[/tex] is 5. In each term of the expansion, the power of [tex]\(x\)[/tex] is given by [tex]\(n - k\)[/tex], so we set [tex]\(n - k = 5\)[/tex]. Therefore, [tex]\(k = 3\)[/tex].
Use the binomial theorem formula for the [tex]\(k\)[/tex]-th term:
[tex]\[ \binom{8}{3} (2x)^{8-3} (3y)^3 \][/tex]
Calculate the binomial coefficient:
[tex]\[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \][/tex]
Now, substitute the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex], and the powers:
[tex]\[ 56 \cdot (2x)^5 \cdot (3y)^3 = 56 \cdot 32x^5 \cdot 27y^3 = 56 \cdot 864x^5y^3 = 48384x^5y^3 \][/tex]
Thus, the term that contains [tex]\(x^5\)[/tex] in the expansion of [tex]\((2x + 3y)^8\)[/tex] is:
[tex]\[ 48384x^5y^3 \][/tex]
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