Question

find the 6th term in the expansion of $(a + b)^5$ in simplest form.
answer attempt 1 out of 2

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 terms are numbered starting from \(k = 0\). For the expansion of \((a + b)^5\), we want to find the 6th term. Since the first term corresponds to \(k=0\), the 6th term corresponds to \(k = 5\).

Step2: Substitute into the binomial formula

For \(n = 5\), \(k = 5\), \(x=a\), and \(y = b\), we have:
\(\binom{5}{5}a^{5-5}b^{5}\)

Step3: Calculate the binomial coefficient and simplify the exponents

First, calculate \(\binom{5}{5}\). Using the formula \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\), we get \(\binom{5}{5}=\frac{5!}{5!(5 - 5)!}=\frac{5!}{5!0!}\). Since \(0!=1\), this simplifies to \(1\). Then, \(a^{5 - 5}=a^{0}=1\) (since any non - zero number to the power of 0 is 1). So the term becomes \(1\times1\times b^{5}=b^{5}\).

Answer:

\(b^{5}\)