Question

cho A=1+4+4^2+4^3+........+4^99 va B=4^100.Chứng tỏ A <\(\frac{1}{3}B\)​

Answer 1

\(=>4A=4+4^2+...+4^{99}+4^{100}\)

\(=>4A-A=\left(4+4^2+...+4^{99}+4^{100}\right)-\left(1+4+4^2+...+4^{99}\right)\)

\(=>3A=4^{100}-1\)

\(=>A=\frac{4^{100}-1}{3}\)

\(\frac{1}{3}B=\frac{4^{100}}{3}\)

=> A<\(\frac{1}{3}B\)

Answer 2

A = 1 + 4 + 4² + 4³ + ... + 4⁹⁹

4A = 4( 1 + 4 + 4² + 4³ + ... + 4⁹⁹ )

4A = 4 + 4² + 4³ + ... + 4¹⁰⁰

4A - A = 3A

= ( 4 + 4² + 4³ + ... + 4¹⁰⁰ ) - ( 1 + 4 + 4² + 4³ + ... + 4⁹⁹ )

= 4 + 4² + 4³ + ... + 4¹⁰⁰ - 1 - 4 - 4² - 4³ - ... - 4⁹⁹

= 4¹⁰⁰ - 1

=> \(A=\frac{4^{100}-1}{3}\)

B = 4¹⁰⁰ => \(\frac{1}{3}B=4^{100}\cdot\frac{1}{3}=\frac{4^{100}}{3}\)

\(4^{100}-1< 4^{100}\Rightarrow\frac{4^{100}-1}{3}< \frac{4^{100}}{3}\Rightarrow A< \frac{1}{3}B\left(đpcm\right)\)