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\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}-\frac{1}{1.3}-\frac{1}{3.5}-\frac{1}{5.7}-...-\frac{1}{11.13}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}-\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(=1-\frac{1}{10}-\frac{1}{2}.\left(1-\frac{1}{13}\right)=\frac{9}{10}-\frac{6}{13}=\frac{57}{130}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{90}-\frac{1}{3}-\frac{1}{15}-.....-\frac{1}{143}\)
\(=\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{90}\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+.....+\frac{1}{143}\right)\)
\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{9.10}\right)-\left(\frac{1}{1.3}+\frac{1}{3.5}+.....+\frac{1}{11.13}\right)\)
\(=\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-.....-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)-\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)\(=\left(\frac{1}{1}-\frac{1}{10}\right)-\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{13}\right)=\frac{9}{10}-\frac{6}{13}=\frac{117}{130}-\frac{78}{130}=\frac{39}{130}=\frac{3}{10}\)
E=4/3+4/15+4/35+4/63+...+4/399+4/483
=4/1.3+4/3.5+4/5.7+4/7.9+...+4/19.21+4/21.23
=2(2/1.3+2/3.5+2/5.7+...+2/19.21+2/21.23)
=2(1/1-1/3+1/3-1/5+1/5-1/7+...+1/19-1/21+1/21-1/23)
=2(1/1-1/23)
=2(23/23-1/23)
=2.22/23
=44/23
`1/15+1/35+1/63+1/99+1/143`
`=1/[3.5]+1/[5.7]+1/[7.9]+1/[9.11]+1/[11.13]`
`=1/2(2/[3.5]+2/[5.7]+2/[7.9]+2/[9.11]+2/[11.13])`
`=1/2.(1/3-1/5+1/5-1/7+...+1/11-1/13)`
`=1/2.(1/3-1/13)`
`=1/2 . 10/39`
`=5/39`
=\(\frac{1}{3\cdot5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{11\cdot13}=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{1}{2}.\frac{10}{13}=\frac{5}{13}\)
B=1/15+1/35+1/63+1/99+1/143
B=1/3.5+1/5.7+1/7.9+1/9.11+11.13 (khoảng cách từ 3-5;5-7;7-9;9-11;11-13 la 2)
Suy ra B=1/2(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13) (Ta gop -1/5+1/5;-1/7+1/7;-1/9+1/9;-1/11+1/11 bang 0)
B=1/2(1/3-1/43)=1/2.40/129=20/129
\(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)
\(B=\frac{2}{3×5}+\frac{2}{5×7}+\frac{2}{7×9}+...+\frac{2}{19×21}\)
\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{19}-\frac{1}{21}\)
\(B=\frac{1}{3}-\frac{1}{21}\)
\(B=\frac{2}{7}\)
A=\(\frac{1}{3}\)+\(\frac{1}{6}\)+\(\frac{1}{10}\)+\(\frac{1}{15}\)+...+\(\frac{1}{66}\)
A=\(\frac{1}{1\cdot3}\) +\(\frac{1}{2\cdot3}\) +\(\frac{1}{2\cdot5}\)+...+\(\frac{1}{6\cdot11}\)
A=\(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{5}+...+\frac{1}{6}-\frac{1}{11}\)
A=\(\frac{1}{1}-\frac{1}{11}\)
=>A=\(\frac{10}{11}\)
B=\(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)
2B=\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{19\cdot21}\)
2B=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)
2B=\(\frac{1}{3}-\frac{1}{21}\)
2B=\(\frac{2}{7}\)
B=\(\frac{2}{7}:2\)
=>B=\(\frac{1}{7}\)
Gọi dãy là A ta có :
A = 1/3.5 + 1/5.7 + 1/7.9 + 1/9.11 + 1/11.13
A = 1/2 . ( 2/3.5 + 2/5.7 + 2/7.9 + 2/9.11 + 2/11.13 )
A = 1/2 . ( 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + 1/11 - 1/13 )
A = 1/2 . ( 1/3 - 1/13 )
A = 1/2 . 10/39
A = 5/39
\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
=\(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
=\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
=\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)\)
=\(\frac{1}{2}.\frac{10}{39}\)
=\(\frac{5}{39}\)
1/15+1/35+1/63+1/99+1/143
=1/2x(1/15+1/35+1/63+1/99+1/143)
=1/2x(2/3x5+2/5x7+2/7x9+2/9x11+2/11x13)
=1/2x(1/3-1/5+1/7-1/9+1/9-1/11+1/11-1/13)
=1/2x(1/3-1/13)
=1/2x10/39
=5/39
