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:
- \(n - k = 4\)
- \(k = 2\)
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 \(15\).