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Lời giải :
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
ko chép lại đề :
= \(\frac{1}{1}\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ ......... + \(\frac{1}{98}\)- \(\frac{1}{99}\)+ \(\frac{1}{99}\)- \(\frac{1}{100}\)
= \(1-\frac{1}{100}\)
= \(\frac{99}{100}\)
Xin lỗi mk nhầm
đề là:
\(1\cdot2\cdot3\cdot4\cdot...\cdot99999999999+\left(\frac{1}{2}+\frac{2}{1}+0,5-1+3-5\right)\)
1/1x2 + 1/2x3 +1/3x4 + ......+1/98x99+1/99x100
=1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +......+ 1/98 - 1/99 + 1/99 + 1/100
=(1-1/100)+(1/2 - 1/2 ) + ( 1/3 - 1/3 ) + ...... + (1/98 - 1/98 ) + ( 1/99 - 1/99 )
= 100/100 - 1/100 + 0 + 0 +.....+ 0 + 0
=99/100
vậy GTBT = 99/100
=2/2 - 2/3 - 2/4 - .........- 2/38 - 2/39 - 2/40
= 2/2 - 2/40
=1 - 2/40
=38/40
= 19/20
a,Đặt \(A=\frac{1}{1\times4}+\frac{1}{4\times7}+...+\frac{1}{97\times100}\)
\(\Rightarrow3A=\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{3}{97\times100}\)
\(\Rightarrow3A=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)
\(\Rightarrow3A=1-\frac{1}{100}=\frac{99}{100}\)
\(\Rightarrow A=\frac{99}{300}\)
b, \(\frac{1}{2}\times\frac{2}{3}\times...\times\frac{99}{100}=\frac{1\times2\times...\times99}{2\times3\times...\times1000}=\frac{1}{100}\)
c, \(\frac{3}{4}\times\frac{8}{9}\times...\times\frac{99}{100}=\frac{1.3}{2.2}\times\frac{2.4}{3.3}\times...\times\frac{9.11}{10.10}=\frac{1.2.....9}{2.3.....10}\times\frac{3.4.....11}{2.3.....10}=\frac{1}{10}\times\frac{11}{2}=\frac{11}{20}\) (dấu . là dấu nhân)
\(\frac{12}{35}:\frac{35}{25}=\frac{12}{35}.\frac{25}{35}=\frac{12.25}{35.35}=\frac{12.5.5}{7.5.7.5}=\frac{12}{49}\)
\(\frac{9}{22}.\frac{33}{18}=\frac{9.33}{22.18}=\frac{9.3.11}{11.2.9.2}=\frac{3}{4}\)
P = 2 x 3 + 3 x 4 + ...+ 99 x 100
=> 3 x P = 2 x 3 x 3 + 3 x 4 x 3 + ....+ 99 x 100 x 3
3 x P = 2 x 3 x ( 4-1) + 3 x 4 x (5-2) + ...+ 99 x 100 x ( 101 -98)
3 x P = 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5- 2 x 3 x 4 + ...+ 99 x 100 x 101 - 98 x 99 x 101
3 x P = ( 2 x 3 x 4 + 3 x 4 x 5 + ...+ 99 x 100 x 101) - ( 1 x 2 x 3 + 2 x 3 x 4 + ...+ 98 x 99 x 101)
3 x P = 99 x 100 x 101 - 1 x 2 x 3
\(P=\frac{99x100x101-1x2x3}{3}=333298\)
p=2.3+3.4+4.5+5.6+...+99.100
3p=2.3.3+3.4.3+4.5.3+5.6.3+...+99.100.3
3p=2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+5.6.(7-4)+...+99.100.(101-98)
3p=2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+5.6.7-4.5.6+99.100.101-98.99.100
3p=98.99.100-1.2.3
p=\(\frac{98.99.100-1.2.3}{3}=323398\)
Suy ra 2A=2/1x2x3+2/2x3x4+2/3x4x5+......+2/38x39x40
2A=3-1/1x2x3+4-2/2x3x4+5-3/3x4x5+........+40-38/38x39x40
2A=1/1x2-1/2x3+1/2x3-1/3x4+1/4x5-1/5x6+........+1/38x39-1/39x40
2A=1/2-1/1560
2A=780/1560-1/1560
2A=779/1560
A=779/1560:2
A=779/1560x1/2
A=779/3120
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+.......+\frac{1}{38.39.40}\)
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+.........+\frac{2}{38.39.40}\)
\(2A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+....+\frac{40-38}{38.39.40}\)
\(2A=\frac{3}{1.2.3}-\frac{1}{1.2.3}+\frac{4}{2.3.4}-\frac{2}{2.3.4}+\frac{5}{3.4.5}-\frac{3}{3.4.5}+.......+\frac{40}{38.39.40}-\frac{38}{38.39.40}\)
\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+.......+\frac{1}{38.39}-\frac{1}{39.40}\)
\(2A=\frac{1}{1.2}-\frac{1}{39.40}\)
\(2A=\frac{1}{2}-\frac{1}{1560}\)
\(2A=\frac{779}{1560}\)
\(A=\frac{779}{1560}:2\)
\(A=\frac{779}{3120}\)
C=1x2+2x3+3x4 +......+99x100
<=>3xC=1x2x3+2x3x3+.....+99x100x3
<=>3xC=1x2x3+2x3x(4-1)+.....+99x100x(101-98)
<=>3xC=1x2x3-1x2x3+2x3x4-2x3x4+......-98x99x100+99x100x101
<=>3xC=99x100x101 <=> C=99x100x101 :3 <=> C= 333300
\(C=1.2+2.3+3.4+...+99.100\)
=> 3C = 1.2.3 + 2.3.3 + 3.4.3 + ...+ 99.100.3
3C = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + ...+ 99.100.(101-98)
3C = 1.2.3 + 2.3.4 -1.2.3 + 3.4.5 - 2.3.4 + ...+ 99.100.101-98.99.100
\(3C=\left(1.2.3+2.3.4+3.4.5+...+99.100.101\right)-\left(1.2.3+2.3.4+...+98.99.100\right)\)
\(3C=99.100.101\Rightarrow C=\frac{99.100.101}{3}=333300\)
Dấu . là dấu nhân
3C = 3 x C