smc

Since $a=kb$, we can write $(a-b{)}^n$ as $(kb-b{)}^n$. Expanding, the second term is $-k^{n-1}{b}^n\left(\genfrac{}{}{0ex}{}{n}{1}\right)$, and the third term is $k^{n-2}{b}^n\left(\genfrac{}{}{0ex}{}{n}{2}\right)$, so we can write the equation $$-k^{n-1}{b}^n\left(\genfrac{}{}{0ex}{}{n}{1}\right)+k^{n-2}{b}^n\left(\genfrac{}{}{0ex}{}{n}{2}\right)=0$$ 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 $n\ge 2$, $n$ 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 $\boxed{2k+1}$.