Question

Using the Binomial Theorem, what would be the coefficient of the sixth term in the binomial expansion of (a+b)9 ?(1 point)

Answer 1

To find the coefficient of the sixth term in the binomial expansion of \((a + b)^9\) using the Binomial Theorem, we start by recalling the general formula for the \(k\)-th term in the expansion of \((a + b)^n\):

\[
T_k = \binom{n}{k-1} a^{n-(k-1)} b^{k-1}
\]

where \(n\) is the exponent, \(k\) is the term number, and \(\binom{n}{k-1}\) is the binomial coefficient.

In this case, \(n = 9\) and we want the sixth term, which means \(k = 6\).

Using the formula:

\[
T_6 = \binom{9}{6-1} a^{9-(6-1)} b^{6-1}
\]

Calculating the components:

1. \( \binom{9}{6-1} = \binom{9}{5} \)
2. The exponent of \(a\) is \(9 - (6 - 1) = 9 - 5 = 4\).
3. The exponent of \(b\) is \(6 - 1 = 5\).

Now we compute \(\binom{9}{5}\):

\[
\binom{9}{5} = \frac{9!}{5!(9-5)!} = \frac{9!}{5! \cdot 4!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 126
\]

Putting it all together, the sixth term is:

\[
T_6 = \binom{9}{5} a^4 b^5 = 126 a^4 b^5
\]

Thus, the coefficient of the sixth term in the binomial expansion of \((a + b)^9\) is:

\[
\boxed{126}
\]