\(P=\dfrac{2017}{2.3}+\dfrac{2017}{3.4}+.......+\dfrac{2017}{19.20}\)
\(=2017\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{19.20}\right)\)
\(=2017\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{19}-\dfrac{1}{20}\right)\)
\(=2017\left(\dfrac{1}{2}-\dfrac{1}{20}\right)\)
\(=2017.\dfrac{9}{20}\)
\(=907,65\)
\(A=\dfrac{2016^2+1^2}{2016\cdot1}+\dfrac{2015^2+2^2}{2015\cdot1}+\dfrac{2014^2+3^2}{2014\cdot3}+...+\dfrac{1009^2+1008^2}{1009\cdot1008}\)
và \(B=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}\)Tìm A/B
Cho \(T=\dfrac{2}{2^1}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{2017}{2^{2016}}+\dfrac{2018}{2^{2017}}\) . So sánh T và 3
Tính tổng:
\(S=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{99\cdot100\cdot101}\)
\(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\)
Chứng minh rằng :
\(A< \dfrac{1}{2}\)
So sánh: \(\dfrac{2017}{2018}+\dfrac{2018}{2017}\) với 2
So sánh :
A = \(\dfrac{2017^{17}+1}{2017^{16}+1}\) và B = \(\dfrac{2017^{18}+1}{2017^{17}+1}\)
\(\dfrac{X-1}{2019}+\dfrac{X-2}{2018}=\dfrac{X-3}{2017}+\dfrac{X-4}{2016}\)
1. So sánh
A = \(\dfrac{2015}{2018^3}\) + \(\dfrac{2017}{2018^4}\) và B = \(\dfrac{2017}{2018^3}\) + \(\dfrac{2015}{2018^4}\)
B = \(\dfrac{1}{2^2}+\dfrac{1}{2^3}+.......+\dfrac{1}{2^{2017}}+\dfrac{1}{2^{2018}}\)
tính B
