10 mũ 99 + 10 mũ 98 +...+ 10 mũ 1 + 1
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
TL:
10x102x103x104x...x1099:104950
=101+2+3+..+99:104950
=10(99+1)x99:2:104950
=104950:104950
=100
=1
~Hok tốt~
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1.9920và 999910
=(992)10=980110
Vậy 980110<999910 suy ra 9920<999910
Câu 2. 3500và 7300
3500=(35)100=243100
7300=(73)100=343100
Vậy 243100<343100 => 3500<7300
![](https://rs.olm.vn/images/avt/0.png?1311)
B=1+3+32+...+310
3B=3(1+3+32+...+310)
3B=3+32+33+...+311
3B-B=(3+32+33+...+311)-(1+3+32+...+310)
2B=311-1
B=\(\frac{3^{11}-1}{2}\)
B=88573
![](https://rs.olm.vn/images/avt/0.png?1311)
`#3107.101107`
1.
`a,`
\(A=1+3+3^2+3^3+...+3^{2012}\)
`3A = 3 + 3^2 + 3^3 + ... + 3^2013`
`3A - A = (3 + 3^2 + 3^3 + ... + 3^2013) - (1 + 3 + 3^2 + 3^3 + ... + 3^2012)`
`2A = 3 + 3^2 + 3^3 + ... + 3^2013 - 1 - 3 - 3^2 - 3^3 - ... - 3^2012`
`2A = 3^2013 - 1`
`=> A = (3^2013 - 1)/2`
Vậy, `A = (3^2013 - 1)/2`
`b,`
\(B=1+10+10^2+10^3+...+10^{2023}\)
`10B = 10 + 10^2 + 10^3 + ... + 10^2024`
`10 B - B = (10 + 10^2 + 10^3 + ... + 10^2024) - (1 - 10 + 10^2 + 10^3 + ... + 10^2023)`
`9B = 10 + 10^2 + 10^3 + ... + 10^2024 - 1 - 10^2 - 10^3 - ... - 10^2023`
`9B = 10^2024 - 1`
`=> B = (10^2024 - 1)/9`
Vậy, `B = (10^2024 - 1)/9.`
`a)A=1+3+3^2+3^3+...+3^2012`
`=>3A=3+3^2+3^3+...+3^2013`
`=>3A-A=2A=3^2013-1`
`=>A=(3^2013-1)/2`
`b)B=1+10+10^2+...+10^2024`
`=>10B=10+10^2+10^3+....+10^2025`
`=>10B-B=9B=10^2025-10`
`=>B=(10^2025-10)/9`
Đặt \(A=10^{99}+10^{98}+...+10+1\)
\(\Rightarrow10A=10^{100}+10^{99}+...+10^2+10\)
\(\Rightarrow10A-A=10^{100}-1\)
\(\Rightarrow A=\frac{10^{100}-1}{9}\)