A=1/3*5 + 1/5*7 + ....+ 1/99*101

A=1/2(2/3*5 + 2/5*7 + ...+ 2/99*101)

A=1/2[(1/3-1/5)+(1/5-1/7)+...+(1/99-1/101)]

A=1/2(1/3-1/101)

A=1/2 * 98/303

A=49/303

\(A=\frac{1}{5}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)

\(=\frac{1}{3X5}+\frac{1}{5X7}+\frac{1}{7X9}+\frac{1}{9X11}+...+\frac{1}{99X101}\)

\(2A=\frac{2}{3X5}+\frac{2}{5X7}+\frac{2}{7X9}+\frac{2}{9X11}+...+\frac{2}{99X101}\)

\(=\frac{5-3}{3X5}+\frac{7-5}{7X9}+\frac{9-7}{9X7}+\frac{11-9}{9X11}+...+\frac{101-99}{101}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\)

\(=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)

Từ đó ta suy ra:\(A=\frac{98}{303}:2=\frac{49}{303}\)

2A=2/3.5+2/5.7+2/7.9+...+2/99.101=>2A=1/3-1/5+1/5-1/7+...+1/99-1/100=>2A=1/3-1/100=>2A=97/300=>A=97/600

\(\frac{49}{303}\)

tinh nhanh a= 1/15 + 1/35 + 1/65 + 1/99 ... + 1/9999

\(=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+.......+\frac{1}{99\cdot101}=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)

\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+.....+\frac{1}{9999}=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{99.}\)\(\frac{1}{99.101}\)

                                                            \(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\)

                                                              \(=1-\frac{1}{101}=\frac{100}{101}\)

\(S=1:3+1:15+1:35+...+1:9999\)

\(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{9999}\)

\(S=2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)

\(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(2S=1-\frac{1}{101}\)

\(2S=\frac{100}{101}\)

\(S=\frac{100}{101}:2\)

\(S=\frac{50}{101}\)

1/3+1/15+1/35+1/63+1/99+……+1/9999
=1/(1×3)+1/(3×5)+1/(5×7)+1/(7×9)+1/(9×11)+……+1/(99×101)
=1/2(1-1/3)+1/2(1/3-1/5)+1/2(1/5-1/7)+1/2(1/7-1/9)+1/2(1/9-1/11)+……+1/2(1/99-1/101)
=1/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+……+1/99-1/101)
=1/2(1-1/101)
=1/2×(100/101)
=50/101

1/3+1/15+1/35+1/63+1/99+……+1/9999

=1/(1×3)+1/(3×5)+1/(5×7)+1/(7×9)+1/(9×11)+……+1/(99×101)

=1/2(1-1/3)+1/2(1/3-1/5)+1/2(1/5-1/7)+1/2(1/7-1/9)+1/2(1/9-1/11)+……+1/2(1/99-1/101)

=1/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+……+1/99-1/101)

=1/2(1-1/101)

=1/2×(100/101)

=50/101 

\(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{15}-\dfrac{1}{35}-\dfrac{1}{63}-...-\dfrac{1}{9999}\)

\(=\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+...+\dfrac{1}{9999}\right)\)

\(=\dfrac{1}{2}-\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{99.101}\right)\)

\(=\dfrac{1}{2}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{2}-\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{100}{101}\)

\(=\dfrac{1}{2}-\dfrac{50}{101}\)

\(=\dfrac{1}{202}.\)

h nghĩ lại thấy mk ngu v~

siêu tốc

\(2000+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)=2000+\frac{50}{3.101}\)

Ta có: \(2000+\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)

=\(2000+\left(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+...+\frac{1}{99x101}\right)\)

Đặt A=\(\left(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+...+\frac{1}{99x101}\right)\)

=> 2xA =\(\left(\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+...+\frac{2}{99x101}\right)\)

2xA = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)

2xA = \(\frac{1}{3}-\frac{1}{101}\)

2xA = \(\frac{98}{303}\)

A = \(\frac{98}{606}=\frac{49}{303}\)

=> \(2000+\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)=2000+\frac{49}{303}=\frac{606049}{303}\)

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