Violympic toán 6
Các câu hỏi tương tự
\(\dfrac{X-1}{2019}+\dfrac{X-2}{2018}=\dfrac{X-3}{2017}+\dfrac{X-4}{2016}\)
\(A=\dfrac{2016^2+1^2}{2016\cdot1}+\dfrac{2015^2+2^2}{2015\cdot1}+\dfrac{2014^2+3^2}{2014\cdot3}+...+\dfrac{1009^2+1008^2}{1009\cdot1008}\)
và \(B=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}\)Tìm A/B
B = \(\dfrac{1}{2^2}+\dfrac{1}{2^3}+.......+\dfrac{1}{2^{2017}}+\dfrac{1}{2^{2018}}\)
tính B
DX
13 tháng 3 2021 lúc 21:45
Cho tổng \(T=\dfrac{2}{2^1}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{2020}{2^{2019}}+\dfrac{2021}{2^{2020}}\)
So sánh T với 3
Cho A = ( \(\dfrac{1}{2^2}-1\) ) ( \(\dfrac{1}{3^2}-1\) ) ( \(\dfrac{1}{4^2}-1\)) ..... ( \(\dfrac{1}{2016^2}-1\) ) ( \(\dfrac{1}{2017^2}-1\) ) và B = \(-\dfrac{1}{2}\) . Hãy so sánh A và B
\(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\)
Chứng minh rằng :
\(A< \dfrac{1}{2}\)
Tính:
S=\(\dfrac{1+2+2^2+2^3+..............+2^{2016}}{1-2^{2017}}\)
Tính:
P=\(\dfrac{2017}{2\cdot3}+\dfrac{2017}{3\cdot4}+...+\dfrac{2017}{19\cdot20}\)
So sánh: \(\dfrac{2017}{2018}+\dfrac{2018}{2017}\) với 2