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\(10^1.10^2.10^3....10^8\)
\(=10^{1+2+3+...+8}\)
\(=10^{36}\)
\(10^1.10^2.10^3...10^{\infty}=10^{1+2+3+...+\infty}=10^{\infty}\)
Đặt A= 1.2.3+2.3.4+3.4.5+...+101.102.103
=>4A=1.2.3.4+2.3.4.4+3.4.5.4+...+101.102.103.4
=1.2.3.(4-0)+2.3.4.(5-1)+3.4.5.(6-2)+...+101.102.103.(104-100)
=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+101.102.103.104-100.101.102.103
=101.102.103.104-0.1.2.3
=110355024
=>A=110355024:4=27588756
Ta có : \(A=\frac{1}{98.99.100}+\frac{1}{99.100.101}+\frac{1}{100.101.102}+\frac{1}{101.102.103}\)
\(2A=2.\left(\frac{1}{98.99.100}+\frac{1}{99.100.101}+\frac{1}{100.101.102}+\frac{1}{101.102.103}\right)\)
\(2A=\frac{2}{98.99.100}+\frac{2}{99.100.101}+\frac{2}{100.101.102}+\frac{2}{101.102.103}\)
\(2A=\frac{1}{98.99}-\frac{1}{99.100}+\frac{1}{99.100}-\frac{1}{100.101}+\frac{1}{100.101}-\frac{1}{101.102}+\frac{1}{101.102}-\frac{1}{102.103}\)
\(2A=\frac{1}{98.99}-\frac{1}{102.103}\)
\(2A=\frac{1}{9702}-\frac{1}{10506}\)
\(A=\frac{\left(\frac{1}{9702}-\frac{1}{10506}\right)}{2}\)
\(A=\left(\frac{1}{9702}-\frac{1}{10506}\right).\frac{1}{2}\)
\(A=\frac{1}{9702.2}-\frac{1}{10506.2}\)
\(A=\frac{1}{19404}-\frac{1}{21012}\)
\(A=\frac{21012-19404}{19404.21012}\)
\(A=\frac{1608}{19404.21012}\)
\(A=\frac{134}{19404.1751}\)
\(A=\frac{67}{9702.1751}\)
Đặt biểu thức là A
=> \(4A=101.102.103.\left(104-100\right)+.....+222615.222616.222617.\left(222618-222614\right)\)
\(=101.102.103.104-100.101.102.103+......+222615.222616.222617.222618-222614.222615.222616.222617\)
\(=222615.222616.222617.222618\)
\(\Rightarrow A=\frac{222615.222616.222617.222618}{4}\)