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\(b=\frac{2}{2\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{99\times101}\)
\(b=\left(\frac{1}{3}+\frac{1}{3}\right)-\left(\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}\right)-\frac{1}{101}\)
\(b=\frac{2}{3}-\frac{1}{101}=\frac{202}{303}-\frac{3}{303}\)
\(b=\frac{199}{303}\)
6S = 1.3(5 - 1) + 3.5(7 - 1) + 5.7(9 - 3) + ... + 99.101(103 - 97)
6S = 1.3 + 1.3.5 - 1.3.5 + 3.5.7 - 3.5.7 +..... - 97.99.101 + 99.101.103
6S = 3 + 99.101.103
6S = 3 + 1029897
6S = 1029900
S =1029900 : 6
S = 171650
Ta có S=1.(1+2)+3.(3+2)+5.(5+2)+....+99.(99+2)
=1.1+3.3+5.5+....+99.99 +1.2+3.2+5.2+...+99.2
=12+32+52+...+992+2.(1+3+5+....+99 )
=1.(2-1)+3.(4-1)+5.(6-1)+...+99.(100-1)+2.(1+3+5+...+99)
=1.2+3.4+5.6+...+99.100-1-3-5-....-99+2.(1+3+5+...+99)
=1.2+3.4+5.6+...+99.100+(1+3+5+...+99)
Xét 1.2+3.4+5.6+...+99.100 = (2-1).2+(4-1).4+(6-1).6+....+(100-1).100
=2.2+4.4+6.6+100.100-2-4-6-...-100
=22+42+62+...+1002-(2+4+6+...+100)
=22.(12+22+32+...+502)-(100+2).50:2
=22.22100-2550 ( bạn tự làm thêm 12+22+...+1002=22100 nhé )
=85850
Do đó S= 85850-(99+1).50:2=85850-2500=83350
\(B=-\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{-1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{100}{101}=-\dfrac{50}{101}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)
\(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)
\(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}\)
\(=\frac{250}{101}\)
A = 1×3+3×5+5×7+...+ 97×99+99×101
6A= 1×3×6+3×5×6+5×7×6+...+97×99×6+99×101×6
6A= 1×3×(5+1)+3×5×(7-1)+5×7×(9-3)+...+97×99×(101-95)+99×101×(103-97)
6A = 1×3×5-1×3+3×5×7-1×3×5+5×7×9-3×5×7+7×9×11-5×7×9+,,,+97×99×101-95×97×99+99×101×103-97×99×101
6A= 1×3+99×101×103
6A= 1029900
A= 171650
Đặt biểu thức là A
=> \(A=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{101}\right)\)
=> \(A=\frac{5}{2}\left(1-\frac{1}{101}\right)\)
=> \(A=\frac{5}{2}.\frac{100}{101}\)
=> \(A=\frac{250}{101}\)
\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}\)
\(=\frac{250}{101}\)
bai nay sai de
1.3+3.5+5.7+......+99.101
=1-101
=-100
ko biết đúng hay sai