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:
$\text{(A) } \tfrac{1}{2}k(k-1)\quad \text{(B) } \tfrac{1}{2}k(k+1)\quad \text{(C) } 2k-1\quad \text{(D) } 2k\quad \text{(E) } 2k+1$
Since $a=kb$, we can write $(a-b)^n$ as $(kb-b)^n$.
Expanding, the second term is $-k^{n-1}b^{n}{{n}\choose{1}}$, and the third term is $k^{n-2}b^{n}{{n}\choose{2}}$, so we can write the equation
\[-k^{n-1}b^{n}{{n}\choose{1}}+k^{n-2}b^{n}{{n}\choose{2}}=0\]
Simplifying and multiplying by two to remove the denominator, we get
\[-2k^{n-1}b^{n}n+k^{n-2}b^{n}n(n-1)=0\]
Factoring, we get
\[k^{n-2}b^{n}n(-2k+n-1)=0\]
Dividing by $k^{n-2}b^{n}$ gives
\[n(-2k+n-1)=0\]
Since it is given that $n\ge2$, $n$ cannot equal 0, so we can divide by n, which gives
\[-2k+n-1=0\]
Solving for $n$ gives
\[n=2k+1\] so the answer is $\fbox{E}$.