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Ta có: 3A = 3.(1+3+32+33+...+399+3100)
3A = 3+32+33+...+3100+3101
Suy ra: 3A – A = (3+32+33+...+3100+3101)−(1+3+32+33+...+399+3100)
2A = 3101−1
⇒ A = 3101−1
2
Vậy A = 3101−1
2
![](https://rs.olm.vn/images/avt/0.png?1311)
ban tinh A banh bao nhieu roi nhan voi 3
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1.
a.\(A=1+2^1+2^2+2^3+...+2^{2007}\)
\(2A=2+2^2+2^3+....+2^{2008}\)
b. \(A=\left(2+2^2+2^3+...+2^{2008}\right)-\left(1+2^1+2^2+..+2^{2007}\right)\)
\(=2^{2008}-1\) (bạn xem lại đề)
2.
\(A=1+3+3^1+3^2+...+3^7\)
a. \(2A=2+2.3+2.3^2+...+2.3^7\)
b.\(3A=3+3^2+3^3+...+3^8\)
\(2A=3^8-1\)
\(=>A=\dfrac{2^8-1}{2}\)
3
.\(B=1+3+3^2+..+3^{2006}\)
a. \(3B=3+3^2+3^3+...+3^{2007}\)
b. \(3B-B=2^{2007}-1\)
\(B=\dfrac{2^{2007}-1}{2}\)
4.
Sửa: \(C=1+4+4^2+4^3+4^4+4^5+4^6\)
a.\(4C=4+4^2+4^3+4^4+4^5+4^6+4^7\)
b.\(4C-C=4^7-1\)
\(C=\dfrac{4^7-1}{3}\)
5.
\(S=1+2+2^2+2^3+...+2^{2017}\)
\(2S=2+2^2+2^3+2^4+...+2^{2018}\)
\(S=2^{2018}-1\)
4:
a:Sửa đề: C=1+4+4^2+4^3+4^4+4^5+4^6
=>4*C=4+4^2+...+4^7
b: 4*C=4+4^2+...+4^7
C=1+4+...+4^6
=>3C=4^7-1
=>\(C=\dfrac{4^7-1}{3}\)
5:
2S=2+2^2+2^3+...+2^2018
=>2S-S=2^2018-1
=>S=2^2018-1
![](https://rs.olm.vn/images/avt/0.png?1311)
Có 2A = 2-2^2+2^3-2^4+.....+2^1015-2^2016+2^2017
3A = 2A+A = (2-2^2+2^3-2^4+....+2^2015-2^2016+2^2017) + (1-2+2^2-2^3+....+2^2016)
= 2^2017 + 1
=> 3A - 2^2017 = 2^2017 + 1 - 2^2017 = 1
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Ta có công thức tổng quát như sau:
\(A=n^k+n^{k+1}+n^{k+2}+...+n^{k+x}\Rightarrow A=\dfrac{n^{k+x+1}-n^k}{n-1}\)
Áp dụng ta có:
\(A=1+4+4^2+...+4^6=\dfrac{4^7-1}{3}\)
\(\Rightarrow B-3A=4^7-3\cdot\dfrac{4^7-1}{3}=1\)
______
\(A=2^0+2^1+...+2^{2008}=2^{2009}-1\)
\(\Rightarrow B-A=2^{2009}-2^{2009}+1=1\)
_____
\(A=1+3+3^2+....+3^{2006}=\dfrac{3^{2007}-1}{2}\)
\(\Rightarrow B-2A=3^{2007}-2\cdot\dfrac{3^{2007}-1}{2}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) A = 1 + 3 + 32 + ... + 37
3A = 3 ( 1 + 3 + 32 + .. + 37)
3A = 3 + 32 + 33 + ...+ 38
b) Vì 3A = 3 + 32 + 33 + ...+38
2A = 38- 1
A = ( 38-1) : 2 (Điều phải chứng minh)
a) A = 1 + 3 + 32 + ... + 37
3A = 3 ( 1 + 3 + 32 + .. + 37)
3A = 3 + 32 + 33 + ...+ 38
b) Vì 3A = 3 + 32 + 33 + ...+38
2A = 38- 1
A = ( 38-1) : 2 (Điều phải chứng minh)
Mong bạn tick cho mình
![](https://rs.olm.vn/images/avt/0.png?1311)
(3a+1).(3a+2)
Ta có: nếu a là số lẻ thì 3a+1 là số chẵn
⇒(3a+1).(3a+2)⋮2 (thỏa mãn)
Ta có: nếu a là số chẵn thì 3a+2 là số chẵn
⇒(3a+1).(3a+2)⋮2 (thỏa mãn)
Vậy với mọi a thì (3a+1).(3a+2)⋮2
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$A=1+3+3^2+3^3+...+3^{2021}$
$3A=3+3^2+3^3+...+3^{2022}$
$\Rightarrow 3A-A=(3+3^2+3^3+...+3^{2022}) - (1+3+3^2+3^3+...+3^{2021})$
$\Rightarrow 2A=3^{2022}-1$
$\Rightarrow A=\frac{3^{2022}-1}{2}$
$B-A=\frac{3^{2022}}{2}-\frac{3^{2022}-1}{2}=\frac{1}{2}$
1) Tính
\(A=1+3+3^2+...+3^{2006}\)
\(3A=3\times\left(1+3+3^2+...+3^{2006}\right)\)
\(3A=3+3^2+3^3+...+3^{2007}\)
\(3A-A=\left(3+3^2+3^3+...+3^{2007}\right)-\left(1+3+3^2+...+3^{2006}\right)\)
\(2A=3^{2007}-1\)
\(A=\left(3^{2007}-1\right):2\)
A = 1 + 3 + 32 + ...+ 32005+ 32006
3A = 3 + 32 + ...+ 32005 + 32006 + 32007
3A - A = 32007 - 1
2A = 32007 - 1
A = (\(3^{2007}\) - 1): 2
3A = \(\dfrac{3^{2008}-3}{2}\)