1/1.7+1/7.13+1/13.19+......+1/61.67
1/1.2.3+1/2.3.4+1/3.4.5+.......+1/8.9.10
1/1+2 + 1/1+2+3 + 1/1+2+3+4 +.......+ 1/1+2+3+....+99
CẦN GẤP
E = 36 / 1.7 + 36 / 7.13 + 36 / 13.19 + .... + 36 / 94.100
F = 1 / 10 + 1 / 40 + 1 / 88 + .... + 1 / ( 3a+z) ( 3a+ 5)
G = 1 / 2.3 + 2 / 3.5 + 3 / 5.8 + 4 / 8.12 + 5 / 12.17
E = \(\frac{36}{1\cdot7}+\frac{36}{7\cdot13}+...+\frac{36}{94\cdot100}=\frac{36}{6}\left[\frac{1}{1\cdot7}+\frac{1}{7\cdot13}+...+\frac{1}{94\cdot100}\right]\)
\(=6\left[1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+...+\frac{1}{94}-\frac{1}{100}\right]=6\left[1-\frac{1}{100}\right]\)
\(=6\cdot\frac{99}{100}=\frac{297}{50}\)
F = \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left[3a+2\right]\left[3a+5\right]}\)
\(=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{\left[3a+2\right]\left[3a+5\right]}\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{3a+2}-\frac{1}{3a+5}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3a+5}\right]=\frac{1}{6}-\frac{1}{9a+15}\)
G = \(\frac{1}{2\cdot3}+\frac{2}{3\cdot5}+\frac{3}{5\cdot8}+\frac{4}{8\cdot12}+\frac{5}{12\cdot17}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{12}-\frac{1}{17}\)
\(=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)
E=36/1-36/7+36/7-36/13+...+36/94-36/100
=36-36/100=891/25
1,Tính nhanh
A=1/3+1/3^2+1/3^3+...+1/3^2007+1/3^2008
B=1/3+1/3^2+1/3^3+...+1/3^n-1+1/3^n ; n∈N*
2,Tính tổng
a,S=1/1.2.3+1/2.3.4+1/3.4.5+..+1/2006.2007.2008
b,S=1/1.2.3+1/2.3.4+1/3.4.5+..+1/n.(n+1).(n+2); n∈N*
A = \(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)
3A= \(1+\frac{1}{3}+...+\frac{1}{3^{2006}}+\frac{1}{3^{2007}}\)
3A-A= \(1-\frac{1}{3^{2008}}\)
B = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{n-1}}+\frac{1}{3^n}\)
3B = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{n-2}}+\frac{1}{3^{n-1}}\)
3B - B = \(1-\frac{1}{3^n}\)
Ta có :
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)
\(\Leftrightarrow\)\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2006}}+\frac{1}{3^{2007}}\)
\(\Leftrightarrow\)\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2006}}+\frac{1}{3^{2007}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\right)\)
\(\Leftrightarrow\)\(2A=1-\frac{1}{3^{2008}}\)
\(\Leftrightarrow\)\(2A=\frac{3^{2008}-1}{3^{2008}}\)
\(\Leftrightarrow\)\(A=\frac{3^{2008}-1}{3^{2008}}:2\)
\(\Leftrightarrow\)\(A=\frac{3^{2008}-1}{2.3^{2008}}\)
Vậy \(A=\frac{3^{2008}-1}{2.3^{2008}}\)
1/1.7+1/7.13+1/13.19+1/19.25+1/25.31+1 / 31.37 = ?
1/7+1/91+1/247+1/475+1/775+1/1147=? (1)
ta có: (1) <=>: 1/(1.7)+1/(7.13)+1/(13.19)+1/(19.25)+1/(25.31)+1/(31.37)
=1/6.(1-1/7+1/7-1/13+1/13-1/19+1/19-1/25+1/25-1/31+1/31-1/37)
=1/6.(1-1/37)=6/37
1.Tính:a)\(\frac{2}{1.7}+\frac{2}{7.13}+\frac{2}{13.19}+....+\frac{2}{601.607}\)
b)\(S=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
Câu 1 : 1.2+2.3+3.4+...+30.31
Câu 2 : 1.2.3+2.3.4+3.4.5+...+30.31.32
Câu 3 : 1/1.2+1/2.3+1/3.4+...+1/30.31
Câu 4 ; 1/1.3+1/3.5+...+1/99.101
Câu 5 : 1/1.4+1/4.7+...+1/91.94
Câu 6 : 1/1.2.3+1/2.3.4+1/3.4.5+...+1/31.32.33
Câu 7 : 1.1!+2.2!+3.3!+...+10.10!
Rút gọn tổng sau:
a, A = 1/2+1/2^2+1/2^3+....+1/2^20
b, B = 1/3+1/3^2+1/3^3+....+1/3^21
c, C = 1/1.2.3+1/2.3.4+1/3.4.5+......+1/19.20.21
1) Tìm x biết:
\(\left(1-\dfrac{3}{10}-x\right):\left(\dfrac{19}{10}-1-\dfrac{2}{5}\right)+\dfrac{4}{5}=1\)
2) Tính nhanh
a)\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{10.11.12}\)
b)\(\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}.\dfrac{4^2}{4.5}\)
câu b bài 2:
\(\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\)
\(=\dfrac{1}{5}\)
câu a bài 2:
\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{10\cdot11\cdot12}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}-...-\dfrac{1}{12}\)
\(=1-\dfrac{1}{12}=\dfrac{11}{12}\)
P = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 +...+ 1/n(n+1)(n+2)
S = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 +...+ 1/48.49.50 .
tao có:
2p=2/1.2.3+2/2.3.4+...+2/n.n(+1)n(n+2)
2p=3-1/1.2.3+4-2/1.2.3+...+(n+2)-n/n.(n+1).(n+2)
2p=3/1.2.3-1/1.2.3+4/2.3.4-2/2.3.4+...+(n+2)/n.(n+1).(n+2)-n/n.(n+1).(n+2)
2p=1/1.2-1/2.3+1/2.3-1/3.4+...+1/n.(n+1)-1/(n+1).(n+2)
2p=1/1.2-1/(n+1).(n+2)
2p=(n+!).(n+2)-2/(2n+2).(n+2)
suy ra p=(n+1).(n+2)-2/(2n+2).(2n+4)
2s=3-1/1.2.3+4-2/1.2.3+...+50-48/48.49.50
2s=3/1.2.3-1/1.2.3+4/2.3.4-2/2.3.4+...+50/49.50.48-48/48.50.49
2s=1/1.2-1/2.3+1/2.3-1/3.4+...+1/48.49-1/49.50
2s=1/1.2-1/49.50
'2s=1/2-1/2450
2s=1225/2450-1/2450
2s=1224/2450
s=612/1225
\(P=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)1
\(P=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\right)\)
\(P=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(P=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(P=\frac{\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)}{2}\)
S cx tinh giong v
Rút gọn tổng sau:
a, A = 1/2+1/2^2+1/2^3+...+1/2^20
b, B= 1/3+1/3^2+1/3^3+...+1/3^21
c, C= 1/1.2.3+1/2.3.4+1/3.4.5+...+1/19.20.21
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\right)\)
\(A=1-\frac{1}{2^{20}}\)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{21}}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{20}}\)
\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{20}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{21}}\right)\)
\(2B=1-\frac{1}{3^{21}}\)
\(B=\frac{1-\frac{1}{3^{21}}}{2}\)
\(C=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{19\cdot20\cdot21}\)
\(C=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{19\cdot20\cdot21}\right)\)
\(C=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{19\cdot20}-\frac{1}{20\cdot21}\right)\)
\(C=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{20\cdot21}\right)\)
\(C=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{420}\right)\)
\(C=\frac{1}{2}\cdot\frac{209}{420}\)
\(C=\frac{209}{480}\)