AHSME 1969 Problem 16

When $(a-b)^n,n\ge2,ab\ne0$, is expanded by the binomial theorem, it is found that when $a=kb$, where $k$ is a positive integer, the sum of the second and third terms is zero. Then $n$ equals: $\text{(A) } \tfrac{1}{2}k(k-1) \text{(B) } \tfrac{1}{2}k(k+1) \text{(C) } 2k-1 \text{(D) } 2k \text{(E) } 2k+1$