\(S=1+2+2^2+...+2^9\)

\(\Rightarrow2S=2+2^2+2^3+...+2^{10}\)

\(\Rightarrow S=2^{10}-1\)

Lại có \(5.2^8=\left(2^2+1\right).2^8=2^{10}+2^8\)

Vậy \(S< 5.2^8\)

S=1+2+2^2+2^3+...+2^9​

2S=2+2^2+2^3+...+2^9+2^10

2S-S=(2+2^2+2^3+...+2^9+2^10)-(1+2+2^2+2^3+...+2^9)

S=2^10-1

5.2^8=(2^2+1).2^8=(2^2.2^8)+(1.2^8)=2^10+2^8

Vì 2^10-1<2^10+2^8=> S<5.2^8

Vậy S < 5. 2^8

Ta có: \(S=1+2+2^2+2^3+...+2^9\)

\(\Rightarrow2S=2+2^2+2^3+...+2^{10}\)

\(\Rightarrow2S-S=\left(2+2^2+2^3+...+2^{10}\right)-\left(1+2+2^2+...+2^9\right)\)

\(\Rightarrow S=2^{10}-1\)

Mặt khác: \(5.2^8=\left(1+2^2\right).2^8=2^8+2^2.2^8=2^8+2^{10}\)

Vì \(2^{10}-1< 2^8+2^{10}\Rightarrow S< 5.2^8\)

bạn viết sai đề rồi 2^210=2^2010

\(2A=2.\left(1+2+....+2^{2010}\right)\)

\(2A-A=\left(2+2^2+...+2^{2011}\right)-\left(1+2+...+2^{2010}\right)\)

\(A=2^{2011}-1\)

\(B=2^{2011}-1=>A=B\)

\(A=2^2\left(1+2^2\right)+2^6\left(1+2^2\right)+...+2^{18}\left(1+2^2\right)\)

=5(2^2+2^6+...+2^18) chia hết cho 5

3 mũ 150> 2 mũ 200

3¹⁵⁰ > 2²⁰⁰

^{dung 100%}

       S = 1 +  2 + 2² + 2³ +...+ 2⁹

      2S =       2 + 2² + 2³+...+ 2⁹ + 2¹⁰

2S - S =        2¹⁰ - 1

        S =  2¹⁰ - 1

        P = 5.2⁰ = 5 <  7 = 2³ - 1 < 2¹⁰ -1   = S
S > P  

ai nhanh mik tích cho nhé

\(A=1+2+2^2+2^3+...+2^{2021}\)

\(\Rightarrow2A=2+2^2+2^3+...+2^{2022}\)

\(\Rightarrow A=2A-A=2+2^2+...+2^{2022}-1-2-2^2-...-2^{2021}=2^{2022}-1>2^{2021}-1=N\)

\(a=1+2+2^2+...+2^{2021}\\ \Rightarrow2a=2+2^2+2^3+...+2^{2022}\\ \Rightarrow2a-a=\left(2+2^2+2^3+...+2^{2022}\right)-\left(1+2+2^2+...+2^{2021}\right)\\ \Rightarrow a=2^{2022}-1>2^{2021}-1=n\)

a) \(\left(3^4.57-9^2.21\right):3^5\)

\(=\left(3^4.57-3^4.21\right):3^5\)

\(=\left[3^4\left(57-21\right)\right]:3^5\)

\(=3^4.36:3^5\)

\(=3^4.2^2.3^2:3^5\)

\(=3.4\)

\(=12\)

b) Ta có; \(1^3+2^3+...+9^3=2025\)

\(\Leftrightarrow2^3.\left(1^3+2^3+....+9^3\right)=2^3.2025\)

\(\Leftrightarrow2^3+4^5+...+18^3=16200\)

S=4+4²+4³+4⁴+...+4⁹⁹

4S=4²+4³+4⁴+...+4⁹⁹+4¹⁰⁰

4S-S=4¹⁰⁰-1

3S=4¹⁰⁰-1

S=(4¹⁰⁰-1):3 < 6.4⁹⁸

vậy S < 6.4⁹⁸