a: =(-1)+(-1)+...+(-1)=-1011

b: =(-5)+(-5)+...+(-5)=-175

Ta có \(\dfrac{1}{3^3}< \dfrac{1}{2.3.4}=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)\)

\(\dfrac{1}{5^3}< \dfrac{1}{4.5.6}=\dfrac{1}{2}\left(\dfrac{1}{4.5}-\dfrac{1}{5.6}\right)\\ ...\\ \dfrac{1}{2021^3}< \dfrac{1}{2020.2021.2022}=\dfrac{1}{2}\left(\dfrac{1}{2020.2021}-\dfrac{1}{2021.2022}\right)\)

Cộng VTV ta được

\(VT< \dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{2021.2022}\right)=\dfrac{1}{12}-\dfrac{1}{2\left(2021.2022\right)}< \dfrac{1}{12}\)

\(n^3=n.n^2>n\left(n^2-1\right)=\left(n-1\right)n\left(n+1\right)\)

\(\dfrac{1}{n^3}< \dfrac{1}{\left(n-1\right)n\left(n+1\right)}\)

\(\dfrac{1}{\left(n-1\right)n\left(n+1\right)}=\dfrac{1}{2}.\dfrac{n+1-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\right)\)

\(\dfrac{1}{3^3}+\dfrac{1}{5^3}+.......+\dfrac{1}{2009^3}< \dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+.....\dfrac{1}{2008.2009.2010}=\dfrac{1}{2}\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+.........+\dfrac{1}{2008.2009}-\dfrac{1}{2009.2010}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{2009.2010}\right)\)

\(=\dfrac{1}{2}\)

Ta có: C = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ... + 1/2021.2022.2023

=> C = 1/2. (3-1/1.2.3 + 4-2/2.3.4 + 5-3/3.4.5 + ... + 2023-2021/2021.2022.2023

=> C = 1/2. (1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 + ... + 1/2021.2022 - 1/2022.2023)

=> C = 1/2. (1/1.2 - 1/2022.2023)

- Phần còn lại bạn tự tính chứ số to quá

a) \(S=1+2+3+...+2021\)

\(=\left(2021+1\right).2021:2\)

\(=2043231\)

b) \(P=1+3+5+...+2021\)

\(=\left(2021+1\right).[\left(2021-1\right):2+1]:2\)

\(=2022.1011:2\)

\(=1022121\)

1/ \(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

=\(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).0\)

=\(0\)

mink chịu bài này nó rất khó

1:

a: =7/5(40+1/4-25-1/4)-1/2021

=21-1/2021=42440/2021

b: =5/9*9-1*16/25=5-16/25=109/25