a: A=3^2(1^2+2^2+...+10^2)

=9*385

=3465

b: B=2^3(1^3+2^3+...+10^3)

=8*3025

=24200

Ta có

A = \(\dfrac{1+7+7^2+7^3+...+7^{11}}{1+7+7^2+7^3+...+7^{10}}\)

Đặt C = 1 + 7 + 7² + 7³+...+7¹¹

7C = 7 + 7² + 7³ + ... + 7¹¹ + 7¹²

=> 6C = 7¹² - 1

C = \(\dfrac{7^{12}-1}{6}\)

Đặt D = 1 + 7 + 7² + 7³+...+7¹⁰

7D = 7 + 7² + 7³ + ... + 7¹⁰ + 7¹¹

=> 6D = \(7^{11}-1\)

D = \(\dfrac{7^{11}-1}{6}\)

=> A = \(\dfrac{\dfrac{7^{12}-1}{6}}{\dfrac{7^{11}-1}{6}}\)

A = \(\dfrac{7^{12}-1}{6}\) : \(\dfrac{7^{11}-1}{6}\)

A = \(\dfrac{7^{12}-1}{6}.\dfrac{6}{7^{11}-1}\)

A = \(\dfrac{7^{12}-1}{7^{11}-1}\) = 7, 000000003

Lại có:

B = \(\dfrac{1+3+3^2+3^3+...+3^{11}}{1+3+3^2+3^3+...+3^{10}}\)\

Đặt H = \(1+3+3^2+3^3+...+3^{11}\)

3H = \(3+3^2+3^3+...+3^{12}\)

=> 2H = \(3^{12}-1\)

H = \(\dfrac{3^{12}-1}{2}\)

Đặt Q = \(1+3+3^2+3^3+...+3^{10}\)

3Q = \(3+3^2+3^3+...+3^{10}+3^{11}\)

=> 2Q = \(3^{11}-1\)

Q = \(\dfrac{3^{11}-1}{2}\)

=> B = \(\dfrac{\dfrac{3^{12}-1}{2}}{\dfrac{3^{11}-1}{2}}\)

B = \(\dfrac{3^{12}-1}{2}:\dfrac{3^{11}-1}{2}\)

B = \(\dfrac{3^{12}-1}{2}.\dfrac{2}{3^{11}-1}\)

B = \(\dfrac{3^{12}-1}{3^{11}-1}\)

B = 3, 00001129

Vì 7, 000000003 > 3, 00001129

=> A > B

Vậy A > B

\(\Rightarrow3A=3+3^2+3^3+...+3^{11}\\ \Rightarrow3A-A=\left(3+3^2+...+3^{11}\right)-\left(1+3+...+3^{10}\right)\\ \Rightarrow2A=3^{11}-1\\ \Rightarrow2A+1=3^{11}=3^n\\ \Rightarrow n=11\)

\(A=3\left(1+3+3^2+3^3+...+3^9\right)⋮3\)

A=3(1+3+32+33+...+39)⋮3

A=3(1+3+32+33+...+39)⋮3

\(A=1+3+3^2+3^3+...+3^{10}\)

=>\(3A=3\left(1+3+3^2+3^3+...+3^{10}\right)\)

=>\(3A=3+3^2+3^3+3^4+...+3^{11}\)

=>\(3A-A=\left(3+3^2+3^3+3^4+...+3^{11}\right)-\left(1+3+3^2+3^3+...+3^{10}\right)\)

=>\(2A=3^{11}-1\)

=>\(2A+1=3^{11}\)

=>\(n=3^{11}:3=3^{10}\)

Nguyễn Thanh Tùng trả lời rồi