Question

Cho A= 1+4+4^2+4^3+..+4^99

      B= 4^100

Chứng minh A< B/3

 

Answer 1

\(A=1+4+4^2+...+4^{99}\)

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

=>\(4A-A=\left(4+4^2+4^3+...+4^{100}\right)-\left(1+4+4^2+...+4^{98}\right)\)

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

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

Ta có đpcm

Answer 2

4A=4+4²+4³+4⁴+...+4⁹⁹+4¹⁰⁰

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

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

=> A < B/3