A Hot $2021$ Sum

Let $a_k = (-1)^\frac {k(k+1)}2 k$ . Then

$\{ a_k \} = \{ -1 , -2, 3, 4, - 5, -6, 7, 8, -9, -10, 11, 12, \cdots a_k \} \implies a_k = \begin{cases} - k & \text{if } k \bmod 4 = 1, 2 \\ + k & \text{for } k \bmod 4 = 3, 0 \end{cases}$

Let $S_n = \displaystyle \sum_{k=1}^n (-1)^\frac {k(k+1)}2 k$ . Then

$\begin{array} {llll} S_1 = - 1 & S_2 = - 3 & S_3 = 0 & S_4 = 4 \\ S_5 = - 1 & S_6 = - 7 & S_7 = 0 & S_8 = 8 \\ S_9 = - 1 & S_{10} = - 11 & S_{11} = 0 & S_{12} = 12 \\ \cdots & \cdots & \cdots & \cdots \end{array}$

$\implies S_n = \begin{cases} -1 & \text{if } n \bmod 4 = 1 \\ -n-1 & \text{if } n \bmod 4 = 2 \\ 0 & \text{if } n \bmod 4 = 3 \\ n & \text{if } n \bmod 4 = 0 \end{cases}$

Since $2021 \bmod 4 = 1$ , $S_{2021} = \boxed{-1}$ .

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Earlier solution:

$\begin{aligned} S & = \sum_{k=1}^{2021} (-1)^\frac {k(k+1)}2 k \\ & = - 1 - 2 + 3 + 4 - 5 - 6 + 7 + 8 - \cdots - 2021 \\ & = \blue{- \sum_{k=0}^{505} (4k+1)} \red{ - \sum_{k=0}^{504} (4k+2) + \sum_{k=0}^{504} (4k+3)} \blue{ + \sum_{k=0}^{505} 4k} \\ & = \blue{- \sum_{k=0}^{505} 1} \red{ + \sum_{k=0}^{504} 1} \\ & = - 506 + 505 = \boxed{-1} \end{aligned}$

$(-1-2+3+4)+(-5-6+7+8)+\ldots +(-2017-2018+2019+2020)-2021=4\times500-2021=-1$