If $J$ is the $n×n$ matrix of all ones, and $A = (l−b)I +bJ$, then $\det(A) = (l − b)^{n−1}(l + (n − 1)b)$ I am stuck on how to prove this by induction. 

Let $J$ be the $n×n$ matrix of all ones, and let $A = (l−b)I +bJ$. Show that $$\det(A) = (l − b)^{n−1}(l + (n − 1)b).$$

I have shown that it holds for $n=2$, and I'm assuming that it holds for the $n=k$ case, $$(l-b)^{k-1}(a+(k-1)b)$$ but I'm having trouble proving that it holds for the $k+1$ case. Please help. 

If you really have to do this by induction, see Ivan Loh's answer. I'll give an induction-free argument using diagonalization. 
The matrix $J$ has rank $1$. Therefore $0$ is an eigenvalue of multiplicity $n-1$. Hence the remaining eigenvalue is given by the trace: it is $n$. It follows that $J$ is diagonalizable. So there exists an invertible matrix $P$ such that 
$$
PJP^{-1}=\mbox{diagonal}(n,0,\ldots,0)
$$
whence
$$
PAP^{-1}=(l-b)I_n+bPJP^{-1}=\mbox{diagonal}((l-b)+nb,l-b,\ldots,l-b).
$$
Therefore
$$
\det A=\det PAP^{-1}=((l-b)+nb)(l-b)^{n-1}=(l-b)^n+nb(l-b)^{n-1}.
$$
So you have a typo and your expression should be

$$
\det A=(l-b)^{n-1}(l+(n-1)b).
$$

