Question

chứng minh rằng: 1/3 + 1/3 mũ 2 + 1/3 mũ 3 + ... + 1/3 mũ 2021<1/2

Answer 1

Đặt \(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2021}}\)

=>\(3A=1+\dfrac{1}{3}+...+\dfrac{1}{3^{2020}}\)

=>\(3A-A=1+\dfrac{1}{3}+...+\dfrac{1}{3^{2020}}-\dfrac{1}{3}-\dfrac{1}{3^2}-...-\dfrac{1}{3^{2021}}\)

=>\(2A=1-\dfrac{1}{3^{2021}}\)

=>\(A=\dfrac{1}{2}-\dfrac{1}{2\cdot3^{2021}}< \dfrac{1}{2}\)

Answer 2

help me pls

Answer 3

A=31​+321​+...+320211​

=>\(3 A = 1 + \frac{1}{3} + . . . + \frac{1}{3^{2020}}\)

=>\(3 A - A = 1 + \frac{1}{3} + . . . + \frac{1}{3^{2020}} - \frac{1}{3} - \frac{1}{3^{2}} - . . . - \frac{1}{3^{2021}}\)

=>\(2 A = 1 - \frac{1}{3^{2021}}\)

=>\(A = \frac{1}{2} - \frac{1}{2 \cdot 3^{2021}} < \frac{1}{2}\)