Question

question 10
1 pts
find the indicated term of each expansion.
ninth term of ((r - t)^{14})
(\bigcirc) (r^{6}t^{8})
(\bigcirc) (3003r^{5}t^{6})
(\bigcirc) (3003r^{9}t^{9})
(\bigcirc) (3003r^{6}t^{8})

Explanation:

Step1: Recall the binomial theorem

The binomial theorem states that \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\), where \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\). For the binomial expansion \((r - t)^{14}\), we can write it as \((r+(-t))^{14}\), so \(a = r\), \(b=-t\), and \(n = 14\).

The \(k\)-th term (starting from \(k = 0\)) of the binomial expansion is given by \(\binom{n}{k}a^{n - k}b^{k}\). We want the ninth term. Since we start counting from \(k = 0\), the ninth term corresponds to \(k=8\) (because when \(k = 0\), it's the first term, \(k = 1\) second term, \(\cdots\), \(k=8\) ninth term).

Step2: Calculate the binomial coefficient

First, calculate \(\binom{14}{8}\). Using the formula \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\), we have:
\[
\binom{14}{8}=\frac{14!}{8!(14 - 8)!}=\frac{14!}{8!6!}=\frac{14\times13\times12\times11\times10\times9\times8!}{8!\times6\times5\times4\times3\times2\times1}=\frac{14\times13\times12\times11\times10\times9}{6\times5\times4\times3\times2\times1}
\]
Simplify the numerator and denominator:
\[
\frac{14\times13\times12\times11\times10\times9}{720}=\frac{14\times13\times12\times11\times10\times9}{720}=3003
\]

Step3: Find the powers of \(a\) and \(b\)

For \(a = r\), \(n - k=14 - 8 = 6\), so \(a^{n - k}=r^{6}\). For \(b=-t\), \(k = 8\), so \(b^{k}=(-t)^{8}=t^{8}\) (since \((-t)^8=t^8\) because the exponent is even).

Step4: Multiply the binomial coefficient, \(a^{n - k}\), and \(b^{k}\)

The ninth term is \(\binom{14}{8}r^{14 - 8}(-t)^{8}=3003\times r^{6}\times t^{8}=3003r^{6}t^{8}\)

Answer:

\(3003r^{6}t^{8}\) (corresponding to the option "3003\(r^{6}t^{8}\)")