1. Tính tổng sau: \(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+...+\frac{1}{990}\)
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Tính tổng ;
M = \(\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+.....+\frac{1}{946}+\frac{1}{990}\)
HELP ME
\(M=\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{1}{946}+\frac{1}{990}\)
\(\Rightarrow\frac{1}{2}M=\frac{1}{2}\left(\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{1}{946}+\frac{1}{990}\right)\)
\(\Rightarrow\frac{1}{2}M=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{1892}+\frac{1}{1980}\)
\(\Rightarrow\frac{1}{2}M=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{43.44}+\frac{1}{44.45}\)
\(\Rightarrow\frac{1}{2}M=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{43}-\frac{1}{44}+\frac{1}{44}-\frac{1}{45}\)
\(\Rightarrow\frac{1}{2}M=\frac{1}{5}-\frac{1}{45}=\frac{9}{45}-\frac{1}{45}=\frac{8}{45}\)
\(\Rightarrow M=\frac{8}{45}:\frac{1}{2}=\frac{8}{45}.2=\frac{16}{45}\)
nhớ ấn đúng cho mình nha
\(M=\frac{2}{30}+\frac{2}{42}+...+\frac{2}{1980}\)
\(=2\left(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{44.45}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{44}-\frac{1}{45}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{45}\right)\)
\(=2\times\frac{8}{45}\)
\(=\frac{16}{45}\)
Chào bạn, bạn hãy theo dõi bài giải của mình nhé!
\(\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{1}{946}+\frac{1}{990}\)
\(=\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+...+\frac{2}{1892}+\frac{2}{1980}\)
\(=\frac{2}{5\cdot6}+\frac{2}{6\cdot7}+\frac{2}{7\cdot8}+...+\frac{2}{43\cdot44}+\frac{2}{44\cdot45}\)
\(=2\left(\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{43\cdot44}+\frac{1}{44\cdot45}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{43}-\frac{1}{44}+\frac{1}{44}-\frac{1}{45}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{45}\right)=2\left(\frac{9}{45}-\frac{1}{45}\right)=2\cdot\frac{8}{45}=\frac{16}{45}\)
Chúc bạn học tốt!
B = 2 - 4 - 6 + 8 + 10 - 12 -14 + 16 + ...+ 2010 - 2012 - 2014 + 2016
= (2 - 4 - 6 + 8 ) + ( 10 - 12 -14 +16 ) + ...+ ( 2010 - 2012 - 2014 + 2016 )
= 0 + 0 +...+ 0 + 0 (có 252 số hạng 0)
= 0
Ta có: Từ 2 đến 2016 có \(\frac{2016-2}{2}+1=1008\)
B=(2-4)-(6-8)+(10-12)-(14-16)+...+(2010-2012)-(2014-2016) như vậy có 1008:2=504 nhóm
=-2+2-2+2+...-2+2 có 504 số hạng trong đó có: 525 số -2 và 525 số +2
=0
\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{10}}=\frac{1023}{1024}\)
BẤM ĐÚNG NHÉ
Tính nhanh :
\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)
\(A=2\left(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+\frac{1}{8\cdot10}+\frac{1}{10\cdot12}+\frac{1}{12\cdot14}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{14}\right)\)
\(A=2\cdot\frac{3}{7}\)
\(A=\frac{6}{7}\)
\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)
\(A=\frac{2}{8}+\frac{2}{24}+\frac{2}{48}+\frac{2}{80}+\frac{2}{120}+\frac{2}{168}\)
\(A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}+\frac{2}{12.14}\)
\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\)
\(A=\frac{1}{2}-\frac{1}{14}\)
\(A=\frac{3}{7}\)
_Chúc bạn học tốt_
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{19.20.21}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{19.20}-\frac{1}{20.21}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{20.21}\right)\)
\(=\frac{1}{2}.\frac{209}{420}\)
\(=\frac{209}{840}\)
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{19\cdot20\cdot21}\)
\(=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{19\cdot20\cdot21}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{19\cdot20}-\frac{1}{20\cdot21}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot21}\right)\)
bn tự lm tp
\(C=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{998\cdot999\cdot1000}\)
\(C=\frac{1}{2}\left[\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{998\cdot999\cdot1000}\right]\)
\(C=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{998\cdot999}-\frac{1}{999\cdot1000}\right]\)
\(C=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{999\cdot1000}\right]\)
Tính nốt :v
Ta có
\(C=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{998\cdot999\cdot1000}\)
\(\Rightarrow2C=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{998\cdot999\cdot1000}\)
\(=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{998\cdot999}-\frac{1}{999\cdot1000}\)
\(=\frac{1}{1\cdot2}-\frac{1}{999\cdot1000}\)
\(=\frac{1}{2}-\frac{1}{999000}\)
\(=\frac{499500}{999000}-\frac{1}{999000}\)
\(=\frac{499499}{999000}\)
\(\Rightarrow C=\frac{499499}{1998000}\)
đúng nha bạn nhớ k mik
Đặt tổng trên = A
Có : A = 1/1.2.3 + 1/2.3.4 + ...... + 1/9.10.11
2A = 2/1.2.3 + 2/2.3.4 + ...... + 2/9.10.11
= 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ....... + 1/9.10 - 1/10.11
= 1/1.2 - 1/10.11
= 1/2 - 1/110 = 27/55
=> A = 27/55 : 2 = 27/110
Tk mk nha