Problem 105

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 $\fbox{E}$.