BT1: Tính S = \(\frac{1}{1+2}\)+ \(\frac{1}{1+2+3}\)+ \(\frac{1}{1+2+3+4}\)+ ... + \(\frac{1}{1+2+3+...+2018}\)
Hí hí bn nào lm đúg mik tick cho nhe
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
13 tháng 8 2018
Xét TH1 : ( S < 8/9 )
\(\frac{1}{2\cdot2}< \frac{1}{1\cdot2};\frac{1}{3\cdot3}< \frac{1}{2\cdot3};...;\frac{1}{9\cdot9}< \frac{1}{8\cdot9}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}\)
hay \(S< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}\)
\(S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\)
\(S< 1-\frac{1}{9}=\frac{8}{9}\left(1\right)\)
TH2 : ( S > 2/5 )
\(\frac{1}{2\cdot2}>\frac{1}{2\cdot3};\frac{1}{3\cdot3}>\frac{1}{3\cdot4};...;\frac{1}{9\cdot9}>\frac{1}{9\cdot10}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)
hay \(S>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)
\(S>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(S>\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\left(2\right)\)
Từ (1) và (2) => đpcm
Ko tk thì ko phải là ng` nx rồi :)
NV
0
S
19 tháng 11 2018
\(Taco\):
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right).......................\left(1-\frac{1}{1+2+3+.............+2018}\right)\)
\(A=\left(\frac{1+2}{1+2}-\frac{1}{1+2}\right).............\left(\frac{1+2+3+......+2018}{1+2+3+.......+2018}-\frac{1}{1+2+3+......+2018}\right)\)
\(A=\left(\frac{2}{1+2}\right)...........\left(\frac{2+3+.......+2018}{1+2+3+......+2018}\right)\)
\(\Rightarrow A+2017.\left(\frac{1}{3}\right).....\frac{2+3+.....+2018}{1+2+3+...+2018}=1.1.1......1=1\)
\(.................................\)
29 tháng 4 2018
Mk chỉ làm đc bài 2 thôi!
\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)
\(\Rightarrow2S=6+3+\frac{3}{2}+...+\frac{3}{2^8}\)
\(\Rightarrow2S-S=6-\frac{3}{2^9}\)
\(\Rightarrow S=6-\frac{3}{2^9}\)
Chúc bạn học tốt ( sai thì đừng ném đá ) !
29 tháng 4 2018
Ta có :
A = \(\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{50^2}\)< \(\frac{1}{1.1}+\frac{1}{1.2}+...+\frac{1}{49.50}\)
A < \(1-1+1-\frac{1}{2}+...+\frac{1}{49}-\frac{1}{50}\)
A < 1 - 1/50 = 49/50 < 2
Vậy A < 2
16 tháng 4 2018
\(2S=2+1+\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{2017}}\)
\(2S-S=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2017}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{2018}}\right)\)
\(\Rightarrow S=2-\frac{1}{2^{2018}}+1-1+\frac{1}{2}-\frac{1}{2}+.....+\frac{1}{2^{2017}}-\frac{1}{2^{2017}}=2-\frac{1}{2^{2018}}\)\(=\frac{2^{2019}-1}{2^{2018}}\)
3 tháng 5 2019
\(1)\frac{1}{2}x-\frac{3}{5}=\frac{-4}{5}\)
\(\Rightarrow\frac{1}{2}x=\frac{-4}{5}+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}x=\frac{-1}{5}\)
\(\Rightarrow x=\frac{-1}{5}:\frac{1}{2}=\frac{-1}{5}\cdot\frac{2}{1}=\frac{-2}{5}\)
\(\Leftrightarrow x=\frac{-2}{5}\)
\(2)3\frac{1}{5}-2\frac{1}{3}x=-1\frac{3}{5}+1\frac{7}{10}\)
\(\Rightarrow\frac{16}{5}-\frac{7}{3}x=-\frac{8}{5}+\frac{17}{10}\)
\(\Rightarrow\frac{7}{3}x=\frac{16}{5}-\frac{-8}{5}+\frac{17}{10}\)
\(\Rightarrow\frac{7}{3}x=\frac{16}{5}+\frac{8}{5}+\frac{17}{10}\)
\(\Rightarrow\frac{7}{3}x=\frac{24}{5}+\frac{17}{10}\)
\(\Rightarrow\frac{7}{3}x=\frac{48}{10}+\frac{17}{10}\)
Đến đây tìm được rồi nhé
3,4, áp dụng bài 1,2 rồi làm :v
Ta có công thức 1 + 2 + ... + n = n(n+1)/2
\(S=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{2018.2019}{2}}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{2019}\right)=...\)tự tính
\(S=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2018}\)
\(=\frac{1}{2.3:2}+\frac{1}{3.4:2}+...+\frac{1}{2018.2019:2}=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2018.2019}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\right)=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2018}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2019}\right)=2.\frac{2017}{4032}=\frac{2017}{2019}\)
Dung giải hay nhỉ? Lâu nay mới on =))))