\(S=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\) so sánh S với \(\dfrac{1}{2}\)

Bài 7 cho S =\(\dfrac{1}{3}+\dfrac{1}{16}+\dfrac{1}{19}+\dfrac{1}{21}+\dfrac{1}{61}+\dfrac{1}{72}+\dfrac{1}{91}+\dfrac{1}{94}\)

So sánh S với \(\dfrac{3}{5}\)

Câu 10 (1,0 điểm)

Cho S = \(\dfrac{1}{2x3}+\dfrac{1}{4x5}+\dfrac{1}{6x7}+...+\dfrac{1}{2020x2021}+\dfrac{1}{2022x2023}\)

So sánh S với \(\dfrac{1011}{2023}\)

Cho S = \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+....+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) so sánh S và \(\dfrac{1}{5}\) 

Cho S = \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)

So sánh: a, S và \(\dfrac{1}{2}\)

                b, S và 1

S=\(\dfrac{2}{2021+1}+\dfrac{2^2}{2021^2+1}+\dfrac{2^3}{2021^{2^2}}+...+\dfrac{2^{n+1}}{2021^{2^n}+1}+...+\dfrac{2^{2021}}{2021^{2^{2020}}+1}\)so sánh S với \(\dfrac{1}{1010}\)

\(S=\dfrac{2}{2021+1}+\dfrac{2^2}{2021^2+1}+\dfrac{2^3}{2021^{2^2}+1}+...+\dfrac{2^{n+1}}{2021^{2^n}+1}+...+\dfrac{2^{2021}}{2021^{2^{2020}}+1}\)

So sánh S với \(\dfrac{1}{1010}\)

Cho \(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{10^2}.\) Chứng minh rằng: \(S>\dfrac{9}{22}\)

So sánh tổng S với 2⁵¹

S = \(\dfrac{1}{2}-\dfrac{1}{3.7}-\dfrac{1}{7.11}-\dfrac{1}{11.15}-\dfrac{1}{15.19}-\dfrac{1}{19.23}-\dfrac{1}{23.27}\)

Mai mk thi r cho mình xem cách làm bài này nhé. Giúp mình với. HELP ME !!!