Question

Cho A = 1 + 4 + 4² + 4³ + ... + 4⁹⁹    và B = 4¹⁰⁰

Chứng minh rằng A< \(\frac{B}{3}\)

Answer 1

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

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

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

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

Answer 2

A = 1 + 4 + 4^2 + 4^3 + ....+ 4^99 

4A = 4 + 4^2 + 4^3 + ..... + 4^100 

4A - A = ( 4 + 4^2 + 4^3 + ..... + 4^100 ) - ( 1 + 4 + 4^2 + 4^3 + .... + 4^99 )

3A = 4^100 - 1 

A = 4^100 - 1 /3 < 4^100/3 

Vậy A < B/3

Answer 3

dễ ẹc!!!!!!!!