A=3^1+3^2+....3^100 Tính 2a+1
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) A = 1 + 3 + 3^2 + ...+ 3^100
=> 3A = 3 + 3^2 + 3^3 + ...+ 3^101
=> 3A - A = 3101 - 1
2A = 3^101-1
\(A=\frac{3^{101}-1}{2}\)
b) ta co: 2A = 3101 - 1 ( phan a)
=> 2A + 1 = 3101 = 32.50+1
=> n = 50
a. A = 1 + 3 + 3\(^2\) + ..... + 3 \(^{100}\)
\(\Rightarrow\) 3A = 3 + 3\(^2\) + ... + 3 \(^{100}\) + 3 \(^{101}\)
\(\Rightarrow\) 3 A - A = 3\(^{101}\) - 1
\(\Rightarrow\) 2A = 3\(^{101}\) - 1
\(\Rightarrow\) A = \(\frac{3^{101}-1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
A=1/3 - 2/3^2+3/3^3 - 4/3^4+ ... - 100/3^100
=>3A=1 -2/3 +3/3^2 - 4/3^3+ ... - 100/3^99
=>4A=A+3A=1-1/3+1/3^2-1/3^3+...-1/3^99 - 100/3^100
=>12A=3.4A=3-1+1/3-1/3^2+...-1/3^98 - 100/3^99
=>16A=12A+4A=3-1/3^99-100/3^99-100/3^1...
<=>16A=3-101/3^99-100/3^100
<=>A=3/16-(101/3^99+100/3^100)/16 < 3/16
Suy ra A<3/16
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
3A = 3 + 3^2 + 3^3 + .. + 3^100+ 3^101
A = 1 + 3 + 3^2 + .. + 3^100
3A - A = 3 + 3^2 + 3^3 + .. + 3^100 + 3^101 - 1 - 3 - 3^2 - ... - 3^100
= 3^101 - 1
2A = 3^101 - 1
2A + 3 = 3^101 - 1 + 3 = 3^ 101 + 2 khác 3^n
=> không có n thỏa mãn
Ta có: A=1+3+32+…+3100
=>A.3=3+32+33+…+3101
=>A.3-A=3+32+33+…+3101-1-3-32-…-3100
=>A.2=3101-1
=>A.2+1=3101=3n
=>3101=3n
=>n=101
Vậy n=101
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 1 + 3 + 32 + 33 + ... + 399
3A = 3 + 32 + 33 + .. + 3100
3A -A = 3 + 32 + 33 + ... + 3100 - 1 - 3 - 32 - 399
2A = 3100 - 1
B - 2A = 3100 - ( 3100 - 1 ) = 1