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A = \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{129.15}\)
A = \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{129.15}\)
A = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{129}-\frac{1}{15}\)
A = \(\frac{1}{3}-\frac{1}{15}\)= \(\frac{5}{15}-\frac{1}{15}\)
A = \(\frac{4}{15}\)
CẢM ƠN LỜI GỢI Ý
A=1/15 + 1/35 + 1/63 + 1/99 + ... + 1/9999
A= 1/3x5 + 1/5x7 + 1/7x9 + 1/9x11 + ...+ 1/99x101
Ax2= 2/3x5 + 2/5x7 + 2/7x9 + 2/9x11 + ... + 2/99x101
Ax2= 5-3/3x5 + 7-5/5x7 + 9-7/7x9 + 11-9/9x11 + ... + 101-99/99x101
Ax2=5/3x5 - 3/3x5 + 7/5x7 - 5/5x7 + 9/7x9 -7/7x9 + ... + 101/99x101 -99/99x101
Ax2=1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/99 - 1/101
Ax2= 1/3 - 1/101
Ax2 = 98/303
A= 98/303 : 2
A=49/303
Đăt S=1/15+1/35+1/63+1/99+...+1/2915+1/3135
=1/3.5+1/5.7+1/7.9+1/9.11+...+1/53.55+1/55.57
=1/2(2/3.5+2/5.7+2/7.9+...+2/53.55+2/55.57)
=1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/53-1/55+1/55-1/57)
=1/2(1/3-1/57)
=1/2(19/57-1/57)
=1/2.18/57
=3/19
Vậy 1/15+1/35+1/63+1/99+...+1/2915+1/3135=3/19
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Đặt \(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{2915}+\frac{1}{3135}\)
\(\Leftrightarrow A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+....+\frac{1}{53\cdot55}+\frac{1}{55\cdot57}\)
\(\Leftrightarrow2A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{53\cdot55}+\frac{2}{55\cdot57}\)
\(\Leftrightarrow2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-....+\frac{1}{53}-\frac{1}{55}+\frac{1}{55}-\frac{1}{57}\)
\(\Leftrightarrow2A=\frac{1}{3}-\frac{1}{57}=\frac{6}{19}\)
\(\Leftrightarrow A=\frac{6}{19}:2=\frac{3}{19}\)
A = 1/15 + 1/35 + 1/63 + 1/99 + ....... + 1/9999
A = 1/3 x 5 + 1/5 x 7 + 1/9 x7 + .........+ 1/99 x101
A = 1/2 x ( 1/3 -1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ...... + 1/99 - 1/101
A = 1/2 x ( 1/3 - 1/99 )
A = 1/2 x 98/303
A = 49/303
A = 1/3.5 +1/5.7 + 1/7.9 + 1/9.11 + ... + 1/99. 101
= 1/2.(2/3.5+ 2/5.7 + 2/7.9 + ...+2/99.101)
= 1/2.(1/3 - 1/5 - 1/5 - 1/7 - 1/7 - 1/9 + .... +1/99 - 1/101
=1/2.(1/3 - 1/101)
=1/2 .98//303
=49/303
Dấu . là nhân đó nha bạn
\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)
\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{99.101}\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+\frac{1}{2}.\left(\frac{1}{7}-\frac{1}{9}\right)+...+\frac{1}{2}\left(\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\frac{98}{303}\)
\(=\frac{49}{303}\)
Dấu chấm(.) ở cấp hai là dấu nhân (x)
A=1/15+1/35+1/63+1/99+1/143+1/195+1/255+1/323
=> A = 1/3.5 + 1/5.7 + 1/7.9 + 1/9.11 + 1/11.13 + 1/13.15 + 1/15.17 + 1/17.19
=> 2A = 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/17 - 1/19
=> 2A = 1/3 - 1/19
=> 2A = 16/57 => A = 16/57 : 2 = 8/57
=>=> A = 1/3.5 + 1/5.7 + 1/7.9 + 1/9.11 + 1/11.13 + 1/13.15 + 1/15.17 + 1/17.19
=>=> 2A = 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/17 - 1/19
=> 2A = 1/3 - 1/19
=> 2A = 16/57 => A = 16/57 : 2 = 8/57
1/(3x5) + 1/(5x7) + 1/(7x9) + 1/(9x11)+... + 1/(99x101)
(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+...+1/99-1/101) : 2
(1/3 - 1/101) : 2 = 98/303 : 2
49/303
Bạn đưa về dãy tổng
\(\frac{1}{3.5}+\frac{1}{5.7}+.....+\)
Có thể tính nhanh vì đây là dãy đặc biệt