Equal coefficients

In the expansion of $(ax+b)^n$ , the coefficient of $x^k$ is $C_k = \left(\begin{array}{c} n \\ k\end{array}\right) a^k b^{n-k}.$ Two successive coefficients are equal if for some $0 \leq k < n$ , $C_k = C_{k-1} \\ \left(\begin{array}{c} n \\ k\end{array}\right) a^k b^{n-k} = \left(\begin{array}{c} n \\ k-1\end{array}\right) a^{k-1} b^{n-k+1}.$ Divide both sides by $a^k\:b^{n-k}$ and use that $\left(\begin{array}{c} n \\ k\end{array}\right) \div \left(\begin{array}{c} n \\ k-1\end{array}\right) = (n-k+1)/k$ , then we have $\frac{n-k+1}k = \frac b a.$ Add 1 to each side of the equation: $\frac{n+1}k = \frac{a+b}a$

In this case, $\frac{n+1}k = \frac{2017}{1008}.$ Since the fraction on the right is already in lowest terms, the fraction on the left must be identical, with $n + 1 = 2017,\ \ k = 1008.$ Thus we conclude that $n = \boxed{2016}$ . Blessed New Year to you all!