Question

5 . Cho A = 1 + 4 + 4^2 + 4^3 + .............. +4^99 và B = 4^100 

CHỨNG MINH RẰNG : A bé hơn B phần 3

 

Answer 1

Bài làm:

Ta có: \(A=1+4+4^2+4^3+...+4^{99}\)

\(\Rightarrow4A=4+4^2+4^3+4^4+...+4^{100}\)

\(\Rightarrow4A-A=\left(4+4^2+...+4^{100}\right)-\left(1+4+...+4^{99}\right)\)

\(\Leftrightarrow3A=4^{100}-1\)

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

\(\Leftrightarrow A< \frac{B}{3}\)

Answer 2

A = 1 + 4 + 4² + 4³ + ... + 4⁹⁹

4A = 4( 1 + 4 + 4² + 4³ + ... + 4⁹⁹ )

     = 4 + 4² + 4³ + 4⁴ + ... + 4¹⁰⁰

4A - A = 3A

           = ( 4 + 4² + 4³ + 4⁴ + ... + 4¹⁰⁰ ) - ( 1 + 4 + 4² + 4³ + ... + 4⁹⁹ )

           =  4 + 4² + 4³ + 4⁴ + ... + 4¹⁰⁰ - 1 - 4 - 4² - 4³ - ... - 4⁹⁹

           = 4¹⁰⁰ - 1 

3A = 4¹⁰⁰ - 1 => A = \(\frac{4^{100}-1}{3}\)

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

\(4^{100}-1< 4^{100}\Rightarrow\frac{4^{100}-1}{3}< \frac{4^{100}}{3}\)

\(\Rightarrow A< \frac{B}{3}\left(đpcm\right)\)

Answer 3

\(4A=4+4^2+...+4^{100}\) 

\(\Rightarrow4A-A=3A=4^{100}-1\)

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