Question

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

Answer 1

The coefficient is `n`

Explanation:

The binomial theorem tells us that:

`(a+b)^n = sum_(k=0)^n ((n),(k)) 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:

`(a+b)^n = overbrace((a+b)(a+b)...(a+b))^"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.