A Serious Series

The sum can be expressed as $\sum_{k=2}^{2021} \dfrac k{2^{2022 - k}} = \dfrac1{2^{2022}} \sum_{k=2}^{2021} k \cdot 2^k$ By mathematical induction , we can get $\sum \limits_{k=2}^n k \cdot 2^k = (n-1) 2^{n+1}$ . Hence, the sum in question is $\dfrac{2021 - 1}{2^{2022}} \cdot 2^{2022} = \boxed{2020} .$

$\begin{aligned} S & = \frac 2{2^{2020}} + \frac 3{2^{2019}} + \frac 4{2^{2018}} + \cdots + \frac {2021}2 \\ & = \frac {2\cdot 2 + 3 \cdot 2^2 + 4 \cdot 2^3 + \cdots + 2021 \cdot 2^{2020}}{2^{2021}} \\ & = \frac {1 + 2x + 3x^2 + 4x^3 + \cdots + 2021 x^{2020}}{2^{2021}} \bigg|_{x = 2} - \frac 1{2^{2021}} \\ & = \frac d{dx} \left[\frac {x + x^2 + x^3 + x^4 + \cdots + x^{2021}}{2^{2021}} \right]_{x = 2} - \frac 1{2^{2021}} - \frac 1{2^{2021}} \\ & = \frac d{dx} \left[\frac {x^{2022}-x}{2^{2021}(x-1)} \right]_{x = 2} - \frac 1{2^{2021}} \\ & = \frac {(2022x^{2021}-1)(x-1) - x^{2022}+x}{2^{2021}(x-1)^2} \bigg|_{x=2} - \frac 1{2^{2021}} \\ & = \frac {2^{2021}(2022-2)-1+2}{2^{2021}} - \frac 1{2^{2021}} \\ & = \boxed{2020} \end{aligned}$