Question

Să se calculeze suma S=1²-2²-3²--2022²-1-2-2-3-...-2021-2022.​

Answer 1

$\it S=1^2+2^2+3^2+\ ...\ +2021^2+2022^2-1\cdot2-2\cdot3-3\cdot4-\ ...\ -2021\cdot2022\\ \\ 1^2-1\cdot2=1-2=-1\\ \\ 2^2-2\cdot3=2(2-3)=-2\\ \\ 3^2=3\cdot4=3(3-4)=-3\\\vdots\\2021^2-2021\cdot2022=2021(2021-2022)=-2021$

Suma devine:

$S = 2022^2-(1+2+3+\ ...\ +2021)=2022^2-\dfrac{2021\cdot2022^{(2}}{2}=2022^2-2021\cdot1011$