Ta có: \(A=4^0+4^1+4^2+...+4^{20}\)

Nhân A với 4 ta có:

\(4A=4\left(4^0+4^1+4^2+...+4^{20}\right)\)

=> \(4A-A=\left(4^1+4^2+4^3+...+4^{21}\right)-\left(4^0+4^1+4^2+...+4^{20}\right)\)

=> \(A\left(4-1\right)=4^{21}-4^0\)

=> \(3A=4^{21}-1\)

=> \(3A+1=4^{21}=\left(4^3\right)^7=64^7>63^7\)

Vậy 3A + 1 > 63^7.

Lời giải:

$A=1+4+4^2+4^3+....+4^{23}$

$4A=4+4^2+4^3+4^4+...+4^{24}$

$\Rightarrow 4A-A=4^{24}-1$

$\Rightarrow 3A+1=4^{24}=(4^3)^8=64^8> 63^7$

4A =4 +4²+4³ +....+4²⁴

3A =4A-A =4²⁴ -1

=>3A + 1 = 4²⁴ = 64⁸ > 63⁷

Vậy 3A +1 > 63⁷

4A=4.(1+4+4^2+4^3+........+4^23)

4A-1=(4+4^2+4^3+4^4+........+4^23+4^24)

-(1+4^1+4^2+4^3+.........+4^23)

=>3A=4^24-1

=3A+1=4^24

Vì 3A+1=4^24=(4^3)^8=64^8>63^7 (Cơ số lớn hơn , số mũ lớn hơn)

Vậy 3A+1>63^7

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Có: A=4⁰+4¹+4²+4³+...+4²⁰

=> 4A = 4+4²+4³+4⁴+........+4²¹

=> 4A-A=4+4²+4³+4⁴+........+4²¹-4⁰-4¹-4²-4³-...-4²⁰

=> 3A= 4²¹-1

=> 3A+1=4²¹-1+1=4²¹=64⁷

=> 3A+1>63⁷

Vậy 3A+1>63⁷

\(A = 1 + 4 + 4^2 + ... + 4\)^\(20\)

\(4A = 4 + 4^2 + 4^3 + ...+ 4\)^\(21\)

\(4A - A = ( 4+ 4^2 + 4^3 + ... + 4\)^\(21\)\()\)\(- ( 1 + 4 + 4^2 + ... + 4\)^\(20\) \()\)

\(3A = 2\)^\(21\) \(- 1\)

\(\Leftrightarrow\)\(3A + 1 = 2\)^\(21\)\(= ( 2^3)^7\)\(= 8^7\)

\(Ta có : 8^7 < 63^7 \)

\(Nên 3A + 1 < 63^7\)

Vì A= 4^0 + 4^1 + 4^2+ 4^3+....+4^20

Suy ra: 4A= 4^1+4^2+4^3+4^4+......+ 4^21

Suy ra:4A-A= 4^21 - 4^0

Suy ra: 3A = 4^21-1

Suy ra: A= (4^21-1) : 3

Suy ra: 3A+1= 3. [ ( 4^21-1) : 3] +1

Suy ra: 3A+1 = ( 4^21-1)+1

Suy ra: 3A + 1 = 4^21= (4^3)^7=64^7

Vì 64 > 63; 7=7

Suy ra: 64^7 > 63^7 hay 3A+1 > 63^7