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3) 2 + 22= 2 + 2.2 = 2 .( 1+2 ) = 2. 3
các phần còn lại tương tụ nhé !
a) Ta có:
A = 1 + 2 + 22 + 23 + ... + 2200
=> 2A = 2(1 + 2 + 22 + 23 + ... + 2200)
=> 2A = 2 + 22 + 23 + 24 + ... + 2201
=> 2A - A = (2 + 22 + 23 + 24 + ... + 2201) - (1 + 2 + 22 + 23 + ... + 2200)
=> A = 2201 - 1
=> A + 1 = 2201 - 1 + 1
=> A + 1 = 2201
Vậy A + 1 = 2201
b) Ta có:
B = 3 + 32 + 33 + ... + 32005
=> 3B = 3(3 + 32 + 33 + ... + 32005)
=> 3B = 32 + 33 + 34 + ... + 32006
=> 3B - B = (32 + 33 + 34 + ... + 32006) - (3 + 32 + 33 + .. + 32005)
=> 2B = 32006 - 3
c) Ta có:
C = 4 + 22 + 23 + ... + 22005
Đặt M = 22 + 23 + ... + 22005, ta có:
2M = 2(22 + 23 + ... + 22005)
=> 2M = 23 + 24 + ... + 22006
=> 2M - M = (23 + 24 + ... + 22006) - (22 + 23 + ... + 22005)
=> M = 22006 - 22
=> M = 22006 - 4
Thay M = 22006 - 4 vào C, ta có:
C = 4 + (22006 - 4) = 22006
=> 2C = 2 . 22006 = 22007
Vậy 2C là lũy thừa của 2.
\(A=\left(3^{60}+3^{58}+3^{56}+...+3^2\right)-\left(3^{59}+3^{57}+3^{55}+...+3\right).\)
\(B=3^{60}+3^{58}+3^{56}+...+3^2\)
\(9B=3^{62}+3^{60}+3^{58}+...+3^4\)
\(B=\frac{9B-B}{8}=\frac{3^{62}-3^2}{8}=\frac{3^2\left(3^{60}-1\right)}{8}\)
\(C=3^{59}+3^{57}+3^{55}+...+3\)
\(9C=3^{61}+3^{59}+3^{57}+...+3^3\)
\(C=\frac{9C-C}{8}=\frac{3^{61}-3}{8}=\frac{3\left(3^{60}-1\right)}{8}\)
\(A=B-C=\frac{3^2\left(3^{60}-1\right)-3\left(3^{60}-1\right)}{8}=\frac{6\left(3^{60}-1\right)}{8}\)
\(A=\frac{2.3.\left(3^{60}-1\right)}{8}=\frac{2.3.3^{60}}{8}-\frac{2.3}{8}=\frac{3^{61}}{4}-\frac{3}{4}=\frac{3^{61}-3}{4}\)
\(A=3+3^2+3^3+....+3^{60}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+....+\left(3^{59}+3^{60}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{59}\left(1+3\right)\)
\(=\left(1+3\right)\left(3+3^3+....+3^{59}\right)\)
\(=4\left(3+3^3+....+3^{59}\right)\)\(⋮\)\(4\)
\(A=3+3^2+3^3+...+3^{60}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+....+\left(3^{58}+3^{59}+3^{60}\right)\)
\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\)
\(=\left(1+3+3^2\right)\left(3+3^4+....+3^{58}\right)\)
\(=13\left(3+3^4+...+3^{58}\right)\)\(⋮\)\(13\)
mà (4;13) = 1
nên A chia hết cho 52
Đặt \(A=2+2^2+2^3+2^4+....+2^{59}+2^{60}\)
\(\Leftrightarrow A=\left(2+2^2\right)+\left(2^3+2^4\right)+.....+\left(2^{59}+2^{60}\right)\)
\(\Leftrightarrow A=2\left(1+2\right)+2^3\left(1+2\right)+....+2^{59}\left(1+2\right)\)
\(\Leftrightarrow A=2\cdot3+2^3\cdot3+....+2^{59}\cdot3\)
\(\Leftrightarrow A=3\cdot\left(2+2^3+....+2^{59}\right)\)
Vậy A chia hết cho 3 (đpcm)
*) Chứng mình A \(⋮\)3
Ta có : A= ( 21 + 22 ) + ( 23 + 24 ) + .... + ( 259 + 260)
= 2. ( 1 + 2 ) + 23 . ( 1 + 2) + ... + 259 . ( 1+ 2)
= 2 . 3 + 23 . 3 + .....+ 259 . 3
= 3. (2 + 23 + .... + 259 ) \(⋮\)3
Vậy A \(⋮\)3 => đpcm
a) A = 22007-1 => A + 1 = 22007
b) Do 2B = 3B - B = 32006- 3 => 2B + 3 = 32006
c) C = 4 + 22 + 23+...+22005 = 22 + 23 + ...+ 22005 + 4
2C - C = 22006 - 22 + 4 =22006 - 22 + 22 = 22006
\(A=1+3+3^2+3^3+...+3^{59}+3^{60}+3^{61}\)
\(=\left(1+3\right)+3^2\left(1+3\right)+...+3^{60}\left(1+3\right)\)
\(=4+3^2.4+...+3^{60}.4\)
\(=4\left(1+3^2+...+3^{60}\right)\)