Question

question
find the 6th term in the expansion of ((a + b)^9) in simplest form.

Explanation:

Step1: Recall the binomial theorem

The binomial theorem states that \((x + y)^n=\sum_{k = 0}^{n}\binom{n}{k}x^{n - k}y^{k}\), where \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\) and the \((k + 1)\)-th term is given by \(\binom{n}{k}x^{n - k}y^{k}\).

Step2: Determine the value of \(k\) for the 6th term

For the 6th term, we have \(k+1 = 6\), so \(k=5\). Here, \(n = 9\), \(x=a\), and \(y = b\).

Step3: Calculate the binomial coefficient \(\binom{9}{5}\)

\(\binom{9}{5}=\frac{9!}{5!(9 - 5)!}=\frac{9!}{5!4!}=\frac{9\times8\times7\times6\times5!}{5!\times4\times3\times2\times1}=\frac{9\times8\times7\times6}{4\times3\times2\times1}\)
\(9\times2\times7 = 126\) (since \(\frac{8}{4}=2\), \(\frac{6}{3\times2\times1}=1\))

Step4: Find the powers of \(a\) and \(b\)

For the term with \(k = 5\), the power of \(a\) is \(n-k=9 - 5=4\) and the power of \(b\) is \(k = 5\).

Step5: Write the 6th term

The 6th term is \(\binom{9}{5}a^{4}b^{5}=126a^{4}b^{5}\)

Answer:

\(126a^{4}b^{5}\)