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\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}}\)
=>\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{99}}\)
=>\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
=>\(A=1-\frac{1}{2^{100}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/2.A=1/22+1/23+...+1/2101
=>1/2A-A=1/2101-1/2
=>-1/2A=1/2101-1/2
A=(1/2101-1/2):(-1/2)=(1/2101-1/2).(-2)
=1-1/2100
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{17^{100}+17^{96}+17^{92}+....+17^4+1}{17^{102}+17^{100}+17^{98}+....+17^2+1}\)
Gọi \(17^{100}+17^{96}+17^{92}+....+17^4+1\) là B
\(B=17^{100}+17^{96}+17^{92}+....+17^4+1\\ 17^4\cdot B=17^{104}+17^{100}+17^{96}+......+17^8+17^4\\ 17^4\cdot B-B=\left(17^{104}+17^{100}+17^{96}+......+17^8+17^4\right)-\left(17^{100}+17^{96}+17^{92}+....+17^4+1\right)\\ B\cdot\left(17^4-1\right)=17^{104}-1\\ B=\dfrac{17^{104}-1}{17^4-1}\)
Gọi \(17^{102}+17^{100}+17^{98}+....+17^2+1\) là C
\(C=17^{102}+17^{100}+17^{98}+....+17^2+1\\ C\cdot17^2=17^{104}+17^{102}+17^{100}+17^{98}+....+17^2\\ C\cdot17^2-C=\left(17^{104}+17^{102}+17^{100}+17^{98}+....+17^2\right)-\left(17^{102}+17^{100}+17^{98}+....+17^2+1\right)\\ C\cdot\left(17^2-1\right)=17^{104}-1\\ C=\dfrac{17^{104}-1}{17^2-1}\)
=>
\(A=B:C\\ A=\dfrac{17^{104}-1}{17^4-1}:\dfrac{17^{104}-1}{17^2-1}\\ A=\dfrac{17^2-1}{17^4-1}\)
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\(A=1+5+5^2+5^3+...+5^{100}\)
\(5A=5+5^2+5^3+...+5^{100}+5^{101}\)
\(5A-A=5+5^2+5^3+...+5^{100}+5^{101}-\left(1+5+5^2+5^3+...+5^{100}\right)\)
\(4A=5^{101}-1\)
\(A=\frac{5^{101}-1}{4}\)
A = 1+5+52+53+...+5100
5A = 5+52+53+54+....+5101
4A = 5A - A = 5101 - 1
=> A = \(\frac{5^{101}-1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1+2^2+...+2^100
2A=2+2^2+2^3+...+2^101
2A=2^101-1
A=(2^101-1):2
\(B=5^1+5^2+...+5^{199}\)
\(\Rightarrow5B=5^2+5^3+...+5^{200}\)
\(\Rightarrow5B-B=\left(5^2+5^3+...+5^{200}\right)-\left(5^1+5^2+...+5^{199}\right)\)
\(\Rightarrow4B=5^{200}-5\)
\(\Rightarrow B=\frac{5^{200}-5}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: \(2A=2^{101}+2^{100}+...+2^2+2\)
\(\Leftrightarrow A=2^{100}-1\)
b: \(3B=3^{101}+3^{100}+...+3^2+3\)
\(\Leftrightarrow2B=3^{100}-1\)
hay \(B=\dfrac{3^{100}-1}{2}\)
c: \(4C=4^{101}+4^{100}+...+4^2+4\)
\(\Leftrightarrow3C=4^{101}-1\)
hay \(C=\dfrac{4^{101}-1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Trả lời
A = 1 + 21 + 22 + ... + 299 + 2100
2A = 2 + 22 + 23 + ... + 2100 + 2101
2A - A = A = ( 2 + 22 + 23 + ... + 2100 + 2101 ) - ( 1 + 21 + 22 + ... + 299 + 2100 )
A = 2101 - 1
\(A=1+2^1+2^2+...+2^{99}+2^{100}\)
\(2A=2+2^2+...+2^{100}+2^{101}\)
Ta có:\(2A-A=\left(2^1+2^2+...+2^{100}\right)-\left(1+2^1+2^2+...+2^{101}\right)\)
\(A=2^{101}-1\)
#hok tốt#
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+......+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)
\(\Leftrightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{99}}\)
=> 2A - A = 1 - \(\frac{1}{2^{100}}\)
<=> A = 1 - \(\frac{1}{2^{100}}\)
\(A=\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}.\)
\(\Rightarrow2A=1+\frac{1}{2^1}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\)
\(\Rightarrow2A-A=1-\frac{1}{2^{100}}\)
\(A=1-\frac{1}{2^{100}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)