Question 16When $(a-b)^n,$ $n\ge2,$ $ab\ne0,$ 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:$\dfrac12k(k-1)$$\dfrac12k(k+1)$$2k-1$$2k$$2k+1$