Question

question
find the 3rd term in the expansion of ((a + b)^{10}) 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 \((k + 1)\)-th term is given by \(\binom{n}{k}x^{n - k}y^{k}\).

Step2: Identify \(n\), \(k\) for the 3rd term

For the 3rd term, \(k+1 = 3\), so \(k=2\). Here, \(n = 10\), \(x=a\), \(y = b\).

Step3: Calculate the binomial coefficient \(\binom{10}{2}\)

\(\binom{10}{2}=\frac{10!}{2!(10 - 2)!}=\frac{10!}{2!8!}=\frac{10\times9\times8!}{2\times1\times8!}=\frac{10\times9}{2\times1}=45\)

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

For \(k = 2\), the power of \(a\) is \(n-k=10 - 2 = 8\) and the power of \(b\) is \(k = 2\).

Step5: Form the 3rd term

The 3rd term is \(\binom{10}{2}a^{8}b^{2}\). Substituting the value of \(\binom{10}{2}=45\), we get \(45a^{8}b^{2}\).

Answer:

\(45a^{8}b^{2}\)