Thu gọn: \(2^{100}\)+\(2^{99}\)+\(2^{98}\)+...+\(2^2\)+\(2\)+\(1\) = ?
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![](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)
A = 2100 - 299 + 298 - 297 +...+ 22 - 2
=> 2A = 2101 - 2100+299 - 298+...+23-22
=> 2A+A= 2101 -2
=> \(A=\frac{2^{101}-2}{3}\)
phần B bn lm tương tự nha!
![](https://rs.olm.vn/images/avt/0.png?1311)
a) A =1+3+32+33+...+3100
3A = 3 + 32+33+...+3101
3A-A=( 3 + 32+33+...+3101)-(1+3+32+33+...+3100)
2A = 3101-1
A = \(\frac{3^{101}-1}{2}\)
Thùy An làm sai rùi
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{101+100+99+98+...+3+2+1}{101-100+99-98+...+3-2+1}\)
\(=\frac{\left(101+1\right).100:2}{\left(101-100\right)+\left(99-98\right)+...+\left(3-2\right)+1}\)
\(=\frac{5050}{1+1+...+1+1}\)(51 chữ số 1)
= \(\frac{5050}{51}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) B = 2100 - 299 + 298 - 297 + ...+ 22 - 2
=> B x 2 = 2101 - 2100 + 299 - 298 + ...23 - 22
=> B x 2 + B = (2101 - 2100 + 299 - 298 + ...23 - 22 ) + (2100 - 299 + 298 - 297 + ...+ 22 - 2)
<=> B x 3 = 2101 - 2 = 2. ( 299 - 1)
=> B = \(\frac{2.\left(2^{99}-1\right)}{3}\)
Phần c) Làm tương tự Lấy C x 3 rồi + với C.
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a) \(A=1+3+3^2+3^3+...+3^{100}\)
\(\Rightarrow 3A=3+3^2+3^3+...+3^{101}\)
Trừ theo vế:
\(\Rightarrow 3A-A=(3+3^2+3^3+..+3^{101})-(1+3+3^2+...+3^{100})\)
\(2A=3^{101}-1\Rightarrow A=\frac{3^{101}-1}{2}\)
b) \(B=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(\Rightarrow 2B=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
Cộng theo vế:
\(\Rightarrow B+2B=2^{201}-2\)
\(\Rightarrow B=\frac{2^{101}-2}{3}\)
c) Ta có:
\(C=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\)
\(\Rightarrow 3C=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3\)
Cộng theo vế:
\(C+3C=(3^{100}-3^{99}+3^{98}-....+3^2-3+1)+(3^{101}-3^{100}+3^{99}-....+3^3-3^2+3)\)
\(4C=3^{101}+1\Rightarrow C=\frac{3^{101}+1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(3A=3+3^2+...+3^{101}\)
\(\Leftrightarrow2A=3^{101}-1\)
hay \(A=\dfrac{3^{101}-1}{2}\)
b: \(2B=2^{101}-2^{100}+...+2^3-2^2\)
\(\Leftrightarrow3B=2^{101}-2\)
hay \(B=\dfrac{2^{101}-2}{3}\)
c: \(3C=3^{101}-3^{100}+....+3^3-3^2+3\)
=>\(4C=3^{101}+1\)
hay \(C=\dfrac{3^{101}+1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A = 2 ^ 100 + 2 ^ 99 + 2 ^ 98 + ... + 2 ^ 2 + 2 ^ 1
2A = 2 ^ 101 + 2 ^ 100 + 2 ^ 99 + ... + 2 ^ 3 + 2 ^ 2
2A - A = ( 2 ^ 101 + 2 ^ 100 + 2 ^ 99 + ... + 2 ^ 3 + 2 ^ 2 )
- ( 2 ^ 100 + 2 ^ 99 + 2 ^ 98 + ... + 2 ^ 2 + 2 ^ 1 )
A = 2 ^ 101 - 2
\(A=2^{100}+2^{99}+2^{98^{ }}+...+2^2+2^1\)
\(2A=2.\left(2^{100}+2^{99}+...+2^1\right)\)
\(2A=2^{101}+2^{100}+...+2^2+2^1\)
\(A=2A-A\)
\(A=2^{101}-2\)
Đặt A = 2 100 + 2 99 + 2 98 + .... + 22 +2 + 1
2A = 2 101 + 2 100 + 2 99 + .... + 2 3 + 2 2 + 2
2A - A = ( 2 101 + 2 100 + 2 99 + .... + 2 3 + 2 2 + 2 )
- ( 2 100 + 2 99 + 2 98 + .... + 22 +2 + 1 )
A = 2 101 - 1
= 2101 + 2100 + 299 + ... + 22 + 2 + 1
=(2101 - 1): 2