Question

so sánh

A=1+3+3^2+........+3^20 và B=3^21-1

Answer 1

\(A=1+3+3^2+....+3^{20}\)

\(\Leftrightarrow3A=3+3^2+...+3^{21}\)

\(\Leftrightarrow3A-A=\left(3+3^2+...+3^{21}\right)-\left(1+3+....+3^{20}\right)\)

\(\Leftrightarrow2A=3^{21}-1\)

\(\Leftrightarrow A=\dfrac{3^{21}-1}{2}\)

Mà \(B=3^{21}-1\)

\(\Leftrightarrow A< B\)

Answer 2

A=1+3+3^2+........+3^20

3A = 3 . ( 1 + 3 + \(3^2+...+3^{20}\))

3A = 3 + \(3^2+3^3+...+3^{21}\)

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

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

=> A = \(\dfrac{3^{21}-1}{2}\)

Vì \(3^{21}-1\) > \(\dfrac{3^{21}-1}{2}\) nên => A < B

Vậy A < B