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

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 \(3S< 64^{12}\)

Ta co:S=4^0+4^1+4^2+...+4^35

=>4S=4^1+4^2+...+4^36

=>4S-S=(4^1+4^2+...+4^36)-(4^0+4^1+...+4^35)

hay 3S=4^36-1

3S=64^12-1<64^12

Vay 3S<64^12

co gi hoi mik de mik lam tiep nhe

bye...

hello

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

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

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

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

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

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

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

Là sao

ban TL làm đúng rồi câu này dễ mà

em nên gõ công thức trực quan để được hỗ trợ tốt nhất nhé

       D =           \(\dfrac{1}{7^2}\) - \(\dfrac{2}{7^3}\) + \(\dfrac{3}{7^4}\) - \(\dfrac{4}{7^5}\) +........+ \(\dfrac{201}{7^{202}}\) -  \(\dfrac{202}{7^{203}}\)

7 \(\times\) D  =  \(\dfrac{1}{7}\) -  \(\dfrac{2}{7^2}\) +  \(\dfrac{3}{7^3}\) - \(\dfrac{4}{7^4}\)  + \(\dfrac{5}{7^5}\) -.......- \(\dfrac{202}{7^{202}}\)

7D +D  =   \(\dfrac{1}{7}\) - \(\dfrac{1}{7^2}\) + \(\dfrac{1}{7^3}\) - \(\dfrac{1}{7^4}\) + \(\dfrac{1}{7^5}\) -.........-\(\dfrac{1}{7^{202}}\) - \(\dfrac{202}{7^{203}}\)

         D = (  \(\dfrac{1}{7}\) - \(\dfrac{1}{7^2}\) + \(\dfrac{1}{7^3}\) - \(\dfrac{1}{7^4}\) + \(\dfrac{1}{7^5}\) -.........-\(\dfrac{1}{7^{202}}\) - \(\dfrac{202}{7^{203}}\)) : 8

Đặt    B =      \(\dfrac{1}{7}\) - \(\dfrac{1}{7^2}\) + \(\dfrac{1}{7^3}\) - \(\dfrac{1}{7^4}\) + \(\dfrac{1}{7^5}\) -........+\(\dfrac{1}{7^{201}}\).-\(\dfrac{1}{7^{202}}\) 

  7   \(\times\) B = 1 - \(\dfrac{1}{7}\)+\(\dfrac{1}{7^2}\) - \(\dfrac{1}{7^3}\) + \(\dfrac{1}{7^4}\) - \(\dfrac{1}{7^5}\) +.........- \(\dfrac{1}{7^{201}}\)

7B + B   =  1 - \(\dfrac{1}{7^{202}}\)

          B   =  ( 1 - \(\dfrac{1}{7^{202}}\)) : 8

         D  =  [ ( 1 - \(\dfrac{1}{7^{202}}\)): 8  - \(\dfrac{202}{7^{203}}\)] : 8 

          D = \(\dfrac{1}{64}\) - \(\dfrac{1}{64.7^{202}}\) - \(\dfrac{202}{7^{203}.8}\) < \(\dfrac{1}{64}\)

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

SAI ROI

TRALOI DUNG NHUNG CACH LAM SAI

A=4(3^2+1)(3^4+1)(3^8+1)...(3^64+1)

2A=8(3^2+1)(3^4+1)(3^8+1)...(3^64+1)

2A=(3^2-1)(3^2+1)(3^4+1)(3^8+1)...(3^64+1)

2A=(3^4-1)(3^4+1)(3^8+1)...(3^64+1)

2A=(3^8-1)(3^8+1)....(3^64+1)

2A=(3^16-1)...(3^64+1)

......

2A=(3^64-1)(3^64+1)

2A=3^128-1

A=(3^128-1)/2

=> A>B

\(A=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(\Leftrightarrow4A=\left(3^2-1\right)\left(3^2+1\right)...\left(3^{64}+1\right)\)

\(\Leftrightarrow4A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(\Leftrightarrow4A=\left(3^8-1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)

\(\Leftrightarrow4A=\left(3^{16}-1\right)\left(3^{16}+1\right)...\left(3^{64}+1\right)\)

\(\Leftrightarrow4A=\left(3^{32}-1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\)

\(\Leftrightarrow4A=\left(3^{64}-1\right)\left(3^{64}+1\right)\Leftrightarrow4A=3^{128}-1\Leftrightarrow A=\frac{3^{128}-1}{4}\)

Ta có \(\frac{3^{128}-1}{4}< 3^{128}-1\Rightarrow A< B\)

Lâm Huyền:Bạn sai đề rồi B phải là 3¹²⁸-1 chứ !

Sorry,mình tính sai.Bạn thay 4A thành 2A va các số 4 thành số 2 nhé