Question

So sánh A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2016}}\)với 1

Answer 1

Ta có: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\)

\(\Rightarrow2A=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^{2016}}\right)\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)

\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)

\(\Rightarrow A=1-\frac{1}{2^{2016}}\)

\(\Rightarrow A<1\left(đpcm\right)\)

Answer 2

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2016}}\)

=>\(2A=1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{2015}}\)

=>\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2016}}\right)\)

=>\(A=1-\frac{1}{2^{2016}}\)

Vậy \(A=1-\frac{1}{2^{2016}}\)

Answer 3

khó nhỉ ?

Answer 4

2A=\(1+\frac{1}{2}+...........+\frac{1}{2^{2015}}\)

\(2A-A=\left(1+\frac{1}{2}+........+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...........+\frac{1}{2^{2016}}\right)\)

\(2A-A=1-\frac{1}{2^{2016}}\)

\(\Rightarrow A=1-\frac{1}{2^{2016}}<1\)

\(\Rightarrow A<1\)

Answer 5

Vậy \(A=1-\frac{1}{2^{2016}}<1\) nhé!

Answer 6

dễ mà Song Thư

Answer 7

mỗi tội viết hơi lâu 

Answer 8

$A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}$

Answer 9

A = 1/2 + 1/2² + 1/2³ + 1/2⁴ + .... + 1/2²⁰¹⁶

2A   =  2(1/2 + 1/2² + 1/2³ + 1/2⁴ + .... + 1/2²⁰¹⁶)

       = 1 + 1/2 + 1/2² + 1/2³ + ... + 1/2²⁰¹⁵

2A - A = (1 + 1/2 + 1/2² + 1/2³ + ... + 1/2²⁰¹⁵) - (1/2 + 1/2² + 1/2³ + 1/2⁴ + .... + 1/2²⁰¹⁶)

A = 1 + 1/2 + 1/2² + 1/2³ + .... + 1/2²⁰¹⁵ - 1/2 - 1/2² - 1/2^{3 -} 1/2⁴ - .... - 1/2²⁰¹⁶

   = 1 - 1/2²⁰¹⁶

ma 1 - 1/2²⁰¹⁶ < 1

nen A < 1 

Vay A < 1

Answer 10

cảm ơn bạn nha, mk đã giải đc bài này rùi.