\(S=\dfrac{2}{2\cdot6}+\dfrac{2}{6\cdot10}+...+\dfrac{2}{96\cdot100}\\ =\dfrac{1}{2}\left(\dfrac{4}{2\cdot6}+\dfrac{4}{6\cdot10}+...+\dfrac{4}{96\cdot100}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{10}+...+\dfrac{1}{96}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{100}\right)=\dfrac{1}{2}\cdot\dfrac{49}{100}\\ =\dfrac{49}{100}\)
Tính tổng
S=\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{23.25}\)
1.rut gon phan so sau
a. \(\dfrac{-63}{81}\),\(\dfrac{5.6}{9.35}\),\(\dfrac{7.2+8}{2.14.5}\),\(\dfrac{9}{33-6}\),8
b.\(\dfrac{-270}{360}\),\(\dfrac{7}{10^2+6.10^2}\)
\(S=\dfrac{2}{2021+1}+\dfrac{2^2}{2021^2+1}+\dfrac{2^3}{2021^{2^2}+1}+...+\dfrac{2^{n+1}}{2021^{2^n}+1}+...+\dfrac{2^{2021}}{2021^{2^{2020}}+1}\)
So sánh S với \(\dfrac{1}{1010}\)
tính tổng
S=\(\dfrac{5^2}{1.6}\)+\(\dfrac{5^2}{6.11}\)+\(\dfrac{5^2}{11.16}\)+.............+\(\dfrac{5^2}{96.101}\)
Cho S=\(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+\dfrac{4}{4^4}+...+\dfrac{2023}{4^{2023}}\). Chứng minh S < \(\dfrac{1}{2}\)
1/Tính:
a. S=\(\dfrac{5^2}{1.6}\) + \(\dfrac{5^2}{6.11}\)+ \(\dfrac{5^2}{11.16}\) + \(\dfrac{5^2}{16.21}\) + \(\dfrac{5^2}{21.26}\)
b. (1 - \(\dfrac{1}{2}\)) . (1 - \(\dfrac{1}{3}\) ) . (1- \(\dfrac{1}{4}\) ) . ( 1 - \(\dfrac{1}{5}\) ) .... ( 1 - \(\dfrac{1}{19}\) ) . ( 1 - \(\dfrac{1}{20}\))
Mk cần gấp lắm ~help me please~
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 S = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}\)và P = \(\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\). Tính \(\dfrac{S}{P}\)
Cho S=\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\)
và P=\(\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
TÍNH S/P
