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Ta có : \(A=1+2+2^2+2^3+...+2^{20}\)
\(\Rightarrow\)\(2A=2+2^2+2^3+2^4+...+2^{20}+2^{21}\)
\(\Rightarrow\)\(A=2^{21}-1\)
\(\Rightarrow\)\(A=B\)
Chúc bạn học tốt !
A=1+2+2^2+2^3+...+2^20
2A=2+2^2+2^3+2^4+...+2^21
2A-A=2^21-1
=>A=B
A = 1 + 2 + 22 + ... + 220
2A = 2 + 22 + 23 + ... + 221
2A - A = (2 + 22 + 23 + ... + 221) - (1 + 2 + 22 + ... + 220)
A = 221 - 1 < 221 = B
=> A < B
A = 1 + 2 + 22
+ ... + 220
2A = 2 + 22
+ 23
+ ... + 221
2A - A = (2 + 22
+ 23
+ ... + 221) - (1 + 2 + 22
+ ... + 220)
A = 221
- 1 < 221
= B
=> A < B
k cho mk nha $_$
:D
a) \(A=1+2+2^2+...+2^{63}\)
\(\Rightarrow2A=2.\left(1+2+2^2+...+2^{63}\right)\)
\(\Rightarrow2A=2+2^2+...+2^{64}\)
\(\Rightarrow2A-A=2+2^2+...+2^{64}-\left(1+2+2^2+...+2^{63}\right)\)
\(\Rightarrow A=2+2^2+...+2^{64}-1-2-2^2-...-2^{63}\)
\(\Rightarrow A=2^{64}-1\)
Vì \(2^{64}-1=2^{64}-1\Rightarrow A=B\)
b) \(A=3^4+3^5+...+3^{20}\)
\(\Rightarrow3A=3^5+3^6+...+3^{21}\)
\(\Rightarrow3A-A=3^5+3^6+...+3^{21}-3^4-3^5-...-3^{20}\)
\(\Rightarrow2A=3^{21}-3^4\)
\(\Rightarrow A=\frac{3^{21}-3^4}{2}\)
Mà \(B=\frac{3^{21}-3^4}{2}\Rightarrow A=B\)
a) Ta có: 2003^152>2003^20>199^20
Vậy 2003^152>199^20
b) Ta có: 3^39=(3^13)^3=1594323^3
11^21=(11^7)^3=19487171^3
Vì 1594323^3<19487171^3 nên 3^39<11^21
ĐỀ MÌNH LÀM LÀ
B=\(2^{2010}-1\)
Mà
\(A=1+2+2^2+....+2^{2009}.\)
\(2A=2.\left(1+2+2^2+...+2^{2009}\right)\)
\(2A=2.1+2.2+2.2^2+...+2.2^{2009}\)
\(2A=2+4+2.2^2+...+2.2^{2009}\)
\(2A-A=\left(2+4+8+...+2^{2010}\right)-\left(1+2^1+2^2+...2^{2009}\right)\)
\(1A=2^{2010}-1\)
\(\Rightarrow A=B\)
\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{81}\right)\left(1-\frac{1}{100}\right)\)
\(B=\frac{3}{4}\cdot\frac{8}{9}\cdot...\cdot\frac{80}{81}\cdot\frac{99}{100}\)
\(B=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot...\cdot\frac{8.10}{9.9}\cdot\frac{9.11}{10.10}\)
\(B=\frac{\left(1\cdot2\cdot...\cdot8\cdot9\right).\left(3\cdot4\cdot...\cdot10\cdot11\right)}{\left(2\cdot3\cdot..\cdot9\cdot10\right).\left(2\cdot3\cdot...\cdot9\cdot10\right)}\)
\(B=\frac{1\cdot2\cdot...\cdot8\cdot9}{2\cdot3\cdot...\cdot9\cdot10}\cdot\frac{3\cdot4\cdot...\cdot10\cdot11}{2\cdot3\cdot...\cdot9\cdot10}\)
\(B=\frac{1}{10}\cdot\frac{11}{2}=\frac{11}{20}\)
Vì 20 < 21 nên 11/20 > 11/21
Vậy .....
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1,2 dễ ko làm
3,
S = 1 + 2 + 22 + 23 + ... + 29
2S = 2 + 22 + 23 + 24 + ... + 210
2S - S = ( 2 + 22 + 23 + 24 + ... + 210 ) - ( 1 + 2 + 22 + 23 + ... + 29 )
S = 210 - 1
Mà 5 . 28 = ( 1 + 22 ) . 28 = 28 + 210 > 210 > 210 - 1
Vậy S < 5 . 28
P = 1 + 3 + 32 + 33 + ... + 320
3P = 3 + 32 + 33 + 34 + ... + 321
3P - P = ( 3 + 32 + 33 + 34 + ... + 321 ) - ( 1 + 3 + 32 + 33 + ... + 320 )
2P = 321 - 1
P = ( 321 - 1 ) : 2 < 321
Vậy P < 321