# How do you find the coefficient of a^(n-1)*b in the expansion of (a+b)^n?

Dec 3, 2016

The coefficient is $n$

#### Explanation:

The binomial theorem tells us that:

${\left(a + b\right)}^{n} = {\sum}_{k = 0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) {a}^{n - k} {b}^{k}$

where ((n),(k)) = (n!)/((n-k)! k!)

The term in ${a}^{n - 1} . b$ is the one for $k = 1$, with coefficient:

((n),(1)) = (n!)/((n-1)! 1!) = n

Alternatively, consider the product:

${\left(a + b\right)}^{n} = {\overbrace{\left(a + b\right) \left(a + b\right) \ldots \left(a + b\right)}}^{\text{n times}}$

If we multiplied out the right hand side, then the only way we can get terms which contribute to the coefficient of ${a}^{n - 1} b$ is by picking the left hand $a$ term of $n - 1$ of the binomials and one right hand $b$ term.

We can do that in precisely $n$ different ways, each contributing $1$ to the coefficient.