\(4^0+4^1+4^2+4^3+...+4^{35}\\ 4S=4^1+4^2+4^3+4^4+...+4^{36}\\ 4S-S=\left(4^1+4^2+4^3+4^4+...+4^{36}\right)-\left(4^0+4^1+4^2+4^3+...+4^{35}\right)\\ 3S=4^{36}-1=64^{12}-1\\ Vì64^{12}-1< 64^{12}\\ \Rightarrow3S< 64^{12}\)

Ta có: \(64^{12}=\left(4^3\right)^{12}=4^{36}\)

\(S=4^0+4^1+...+4^{34}+4^{35}\)

\(\Rightarrow4S=4^1+4^2+...+4^{35}+4^{36}\)

\(\Rightarrow4S-S=4^{36}-4^0\)

\(\Rightarrow3S=4^{36}-1< 4^{36}\)

Vậy \(3S< 64^{12}\)

4S=4.(4⁰+4¹+4³+...+4³⁵)

4S=4¹+4²+...+4³⁶

4S-S=(4¹-4¹)+(4^2-4²)+...+(3³⁵-3³⁵)+3³⁶-3⁰

3S=0+0+...+0+3³⁶-1

64¹²=(3⁴)¹²=3³⁶

vỉ 3³⁶-1<3³⁶ nên 3S<64¹²

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\(S=4^0+4^1+4^2+4^3+...+4^{35}\)

\(4S=4^1+4^2+4^3+...+4^{36}\)

\(4S-S=(4^1+4^2+4^3+...+4^{36})-(4^0+4^1+4^2+4^3+...+4^{35})\)

\(3S=4^{36}-4^0\)

\(S=4^{36}-1\)

\(\text{Ta thấy :}64^{12}=(4^3)^{12}=4^{36}\)

\(\text{Mà }4^{36}-1>4^{36}\text{ nên }3S>A\)

Là sao

ban TL làm đúng rồi câu này dễ mà

\(S=1+4+4^2+.....+4^{35}\)

\(\Leftrightarrow4S=4+4^2+4^3+........+4^{36}\)

\(\Leftrightarrow4S-S=\left(4+4^2+......+4^{36}\right)-\left(1+4+4^2+......+4^{35}\right)\)

\(\Leftrightarrow3S=4^{36}-1\)

\(\Leftrightarrow3S+1=4^{36}=\left(4^3\right)^9=64^9< 64^{12}\)

\(\Leftrightarrow3S+1< 64^{12}\)

Ta có

S=4⁰+4¹+4²+...+4³⁴+4³⁵

=>4S=4¹+4²+4³+...+4³⁵+4³⁶

=> 4S-S=(4⁰+4¹+4²+...+4³⁴+4³⁵)- (4¹+4²+4³+...+4³⁵+4³⁶)

=> 3S=4³⁶-4⁰=4³⁶-1=64¹²-1

=> 3S<64¹²

Dễ thấy:64^{12}=\left(4^3\right)^{12}=4^{3.12}=4^{36}6412=(43)12=43.12=436
Ta có: 4S=4\left(4^0+4^1+4^2+4^3+...+4^{35}\right)4(40+41+42+43+...+435)
=4^1+4^2+4^3+4^4+...+4^{36}=41+42+43+44+...+436
=>4S-S=4^{36}-4^0436−40
Hay 3S=4^{36}-1< 4^{36}=64^{12}436−1<436=6412
Vậy 3S<64^{12}6412

\(S = 1 + 4 + 4^ 2 + ... + 4\)³⁵

\(4S = 4 + 4^2 + 4 ^ 3 + ... + 4\)³⁶

\(4S - S = ( 1 + 4 + 4^ 2 + ... + \)4³⁶\()\) \(- ( 1 + 4 + 4 ^ 2 + ... + 4\)³⁵ \()\)

\(3S = 4\)³⁶ \(- 1\)

\(3S = 64\)¹² - 11

\(Ta thấy : 64\)¹² \(- 1 < 64\)¹²

\(Do đó : 3S < 64\)¹²

\(Vậy : 3S < 64\)¹²

 S = 4⁰ + 4¹ + 4² + 4³ + ........... + 4³⁵

=> 4S = 4.(   4⁰ + 4¹ + 4² + 4³ + ........... + 4^{35 )}

=> 4S = 4¹+4² + 4³ + ... + 4³⁶

=> 3S = ( 4¹+4² + 4³ + ... + 4³⁶ ) - (  4⁰ + 4¹ + 4² + 4³ + ........... + 4³⁵ )

=> 3S = 4³⁶ - 4⁰ = 4³⁶ - 1

Ta có : 4³⁶ - 1 = ( 4³ )¹² - 1 = 64¹² - 1 < 64¹²

Vậy  3S < 64¹²

Bạn nhân  4S = 4( 4⁰+4¹+......+4³⁵) = 4¹+4²+4³+......+4³⁶

Lấy 4S - S = 3S = 4¹+4²+4³+......+4³⁶- (4⁰+4¹+4²......+4³⁵) = 4³⁶- 1 

3S = 4³⁶- 1 = (4³)¹²-1 = 64¹²-1 < 64¹²

ko biet

Ta có: \(S=4^0+4^1+...+4^{35}\)

\(\Rightarrow4S=4+4^1+...+4^{36}\)

\(\Rightarrow4S-S=\left(4+4^1+...+4^{36}\right)-\left(4^0+4^1+...+4^{35}\right)\)

\(\Rightarrow3S=4^{36}-4^0\)

\(\Rightarrow3S=\left(4^3\right)^{12}-1\)

\(\Rightarrow3S=64^{12}-1\)

Vì \(64^{12}-1< 64^{12}\) nên \(3S< 64^{12}\)

Vậy \(3S< 64^{12}\)

Ta có:

Vì nên

Vậy

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Ta có:

Vì nên

Vậy