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\(a.A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(2A-A=1-\frac{1}{2^{99}}\)
\(A=1-\frac{1}{2^{99}}< 1\)
\(b.B=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\right)\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6A=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6A-2A=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4A=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{303}{3^{100}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{203}{3^{100}}< 3\)
\(A< \frac{3}{4}\)
Ủng hộ mk nha ^_^
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(A=1+2+2^2+2^3+...+2^{100}\)
=> \(2A=2+2^2+2^3+2^4+...+2^{101}\)
=> \(A=2A-A=2^{101}-1\)
=> \(A+1=2^{101}\)
b, \(B=3+3^2+3^3+...+3^{2005}\)
\(3A=3^2+3^3+3^4+....+3^{2006}\)
=> \(2A=3A-A=3^{2006}-3\)
=> \(2A+3=3^{2006}\)là lũy thừa của 3
=> Đpcm
a) Ta có: \(A=1+2+2^2+2^3+.....+2^{100}\)
\(\Rightarrow2A=2+2^2+2^3+........+2^{101}\)
Lấy 2A-A ta có:
\(2A-A=\left(2+2^2+2^3+2^4+.....+2^{101}\right)\)\(-\left(1+2+2^2+2^3+.......+2^{100}\right)\)
\(\Rightarrow A=2^{101}-1\)
\(\Rightarrow A+1=2^{101}-1+1\)
\(\Rightarrow A+1=2^{101}\)
b) Ta có: \(B=3+3^2+3^3+.....+3^{2005}\)
\(\Rightarrow3B=3^2+3^3+3^4+.....+3^{2006}\)
\(\Rightarrow3B-B=\left(3^2+3^3+3^4+....+3^{2006}\right)\)\(-\left(3+3^2+3^3+......+3^{2005}\right)\)
\(\Rightarrow2B=3^{2006}-3\)
\(\Rightarrow2B+3=3^{2006}-3+3\)
\(\Rightarrow2B+3=3^{2006}\)
Vậy 2B+3 là lũy thừa của 3 ĐPCM
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1+3+32+...+37
3A=3+32+33+...+38
3A-A=(3+32+33+...+38)-(1+3+32+...+37)
2A=38-1
A=(38-1):2
mà vì chia 2 nên chắc chắn A chia hết cho 2
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Ta có:3.A=1+1/3+1/3^2+...+1/3^97 +1/3^98
=>3.A - A=(1+1/3+1/3^2+...+1/3^98 + 1/3^99)-(1/3+1/3^2 +1/3^3+...+1/3^98+1/3^99)
=>2.A=1-1/3^99
=>A=1/2 -1/3^99.1/2 <1/2
Vậy ... T I C K cho mình với nha
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2014}}\)
\(A=\left(3A-A\right):2\)
\(3A=3+1+\frac{1}{3}+...+\frac{1}{3^{2013}}\)
\(A=\left(3-\frac{1}{3^{2014}}\right):2\)
\(A=\frac{3}{2}-\frac{1}{2.3^{2014}}\)
\(\Rightarrow A<\frac{3}{2}\)