Geoffrey writes out the binomial expansion of (a+b)^6 using the Binomial Theorem. Which of the following can he use to find the coefficient for the a^4b^2 term?

1 answer

To find the coefficient of the term \(a^4b^2\) in the binomial expansion of \((a+b)^6\), Geoffrey can use the Binomial Theorem. The Binomial Theorem states that:

\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k, \]

where \(\binom{n}{k}\) is the binomial coefficient.

In this case, \(n = 6\), and we need to identify the values of \(k\) that will give us \(a^4b^2\).

From the term \(a^{n-k}b^k\), we see that:

  • \(n - k = 4\) (the exponent of \(a\)),
  • \(k = 2\) (the exponent of \(b\)).

We can set up the equations:

  1. \(n - k = 4\)
  2. \(k = 2\)

Substituting \(k = 2\) into the first equation gives: \[ 6 - k = 4 \implies k = 2 \]

Now that we know \(k = 2\), we can find the coefficient for the term \(a^4b^2\) using the binomial coefficient:

\[ \text{Coefficient} = \binom{6}{2}. \]

Thus, Geoffrey can calculate:

\[ \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15. \]

Therefore, the coefficient for the \(a^4b^2\) term in the expansion of \((a + b)^6\) is \(15\).