Let's address each question one by one, incorporating all necessary details to make it clear.
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Problem 29: The fourth term in the expansion of [tex]\((y - 2x)^{12}\)[/tex].
To find specific terms in a binomial expansion, we use the Binomial Theorem, which states:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} \cdot a^{n-k} \cdot b^k
\][/tex]
Here, we have [tex]\(a = y\)[/tex], [tex]\(b = -2x\)[/tex], and [tex]\(n = 12\)[/tex]. The [tex]\(k\)[/tex]-th term in the expansion (starting from [tex]\(k = 0\)[/tex]) will have the form:
[tex]\[
\binom{12}{k} \cdot y^{12-k} \cdot (-2x)^k
\][/tex]
The fourth term (which corresponds to [tex]\(k = 3\)[/tex], because our sequence starts at [tex]\(k = 0\)[/tex]) can be determined by substituting [tex]\(k = 3\)[/tex]:
[tex]\[
\binom{12}{3} \cdot y^{12-3} \cdot (-2x)^3
\][/tex]
Evaluating each part separately:
- [tex]\(\binom{12}{3}\)[/tex] is the binomial coefficient “12 choose 3”:
[tex]\[
\binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12!}{3! \cdot 9!} = \frac{12 \cdot 11 \cdot 10}{3 \cdot 2 \cdot 1} = 220
\][/tex]
- [tex]\(y^{12-3} = y^9\)[/tex]
- [tex]\((-2x)^3 = -8x^3\)[/tex]
Putting it all together, the fourth term is:
[tex]\[
220 \cdot y^9 \cdot (-8x^3) = 220 \cdot -8 \cdot y^9 \cdot x^3 = -1760 y^9 x^3
\][/tex]
Thus, the correct answer is:
(A) [tex]\(-1760 y^9 x^3\)[/tex]
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Problem 30: If the third term in [tex]\((1 + x)^n\)[/tex] is [tex]\(-\frac{1}{8} x^2\)[/tex], what is the value of [tex]\(n\)[/tex]?
In the binomial expansion of [tex]\((1 + x)^n\)[/tex], the general form of the [tex]\(k\)[/tex]-th term is given by:
[tex]\[
\binom{n}{k} \cdot 1^{n-k} \cdot x^k = \binom{n}{k} \cdot x^k
\][/tex]
For the third term, which corresponds to [tex]\(k = 2\)[/tex] (since [tex]\(k\)[/tex] starts at 0), this expression is:
[tex]\[
\binom{n}{2} \cdot x^2
\][/tex]
Given that this term is [tex]\(-\frac{1}{8}x^2\)[/tex], we equate:
[tex]\[
\binom{n}{2} \cdot x^2 = -\frac{1}{8} x^2
\][/tex]
Since [tex]\(x^2\)[/tex] terms on both sides are equal, we focus on the coefficients:
[tex]\[
\binom{n}{2} = -\frac{1}{8}
\][/tex]
Evaluating [tex]\(\binom{n}{2}\)[/tex] using its definition:
[tex]\[
\binom{n}{2} = \frac{n(n-1)}{2}
\][/tex]
Equating this to [tex]\(-\frac{1}{8}\)[/tex]:
[tex]\[
\frac{n(n-1)}{2} = -\frac{1}{8}
\][/tex]
Multiply both sides by 2:
[tex]\[
n(n-1) = -\frac{1}{4}
\][/tex]
This equation cannot hold as we can never get a valid integer [tex]\(n\)[/tex] making [tex]\(n(n-1) = -\frac{1}{4}\)[/tex].
Hence, the correct choice is:
(E) None of these
