Question

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

So sánh A và B

Answer 1

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