Binomial Coefficients and series

Question

Let $p(x)$ be the polynomial given by :

$$p(x) = [(x−2)^{2016}(x+2016)]+[(x−2)^{2015}(x+2015)]+⋯+(x−2)(x+1).$$

What is the sum of the coefficients of $p(x)$ ?

Answer 1

We have that $$p(x) = \sum_{k = 1}^{2016}(x -2)^k(x+k),$$ and in order to find the sum of the coefficients, it suffices to compute $p(1)$: indeed, suppose that we have a polynomial $$f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots a_1x + a_0,$$ then $f(1) = a_n + a_{n-1} + \ldots + a_1 + a_0$.

Note that $$p(1) = \sum_{k = 1}^{2016}(1 -2)^k(1+k) = \sum_{k = 1}^{2016}(-1)^k(1+k).$$ This is equal to $$- \sum_{k = 1}^{1008}(1 + (2k-1)) + \sum_{k = 1}^{1008}(1 + 2k)$$ (do you see why this is indeed true?). Simplifying this sum gives $$- \sum_{k = 1}^{1008}(2k) + \sum_{k = 1}^{1008}(1 + 2k) = - \sum_{k = 1}^{1008}(2k) + \sum_{k = 1}^{1008}(2k) + \sum_{k = 1}^{1008}1 = \sum_{k = 1}^{1008}1 = 1008 \cdot 1 = 1008.$$

$\textbf{EDIT}$: This could have been done much shorter by noting that \begin{align} p(1) &= -2 + 3 - \ldots - 2014 + 2015 - 2016 + 2017\\ &= (-2 + 3) + \ldots + (-2014+2015) + (-2016 + 2017)\\ &= \underbrace{1 + \ldots 1 + 1}_{1008 \text{ times}}\\ &= 1008.\end{align}