Ta có:A= \(1+2+2^2+2^3+...+2^{2010}\)

=> 2A= 2(\(1+2+2^2+2^3+...+2^{2010}\))

=> 2A= 2 +\(2^2+2^3+2^4+...+2^{2011}\)

=> 2A-A= A =(2+ \(2^2+2^3+2^4+...+2^{2011}\)) -( \(1+2+2^2+2^3+...+2^{2010}\))

=> A= \(2^{2011}-1\)

Mà B = \(2^{2011}\)

=> A < B

A = 2 + 2^2 + 2^3 + 2^4 + ... + 2^2010 hay A = 3 + 2^2 + 2^3 + 2^4 + ... + 2^2010 bạn

Đặt : A = 1 + 2 + 2^2 + 2^3 + ... + 2^2016 

=> 2A = 2 + 2^2 + 2^3 + 2^4 + ... + 2^2017

=> 2A - A = (  2 + 2^2 + 2^3 + 2^4 + ... + 2^2017 ) - ( 1 + 2 + 2^2 + 2^3 + ... + 2^2016 )

=> A = 2^2017 - 1

=> A < 2^2017 

Vậy A < 2^2017

Ta đặt A = 1 + 2 + 2² + 2³ + ....+ 2²⁰¹⁶

     => 2A = 2 + 2² + 2³ + ...+2²⁰¹⁷

      => 2A - A = (2+2²+2³+...+2²⁰¹⁷) - (1+2+2²+...+2^{2016 )}

        =>    A      =    2²⁰¹⁷ - 1

Mà 2²⁰¹⁷ - 1 < 2²⁰¹⁷

=> A < 2²⁰¹⁷

Vậy 1 + 2 + 2² + ...+ 2²⁰¹⁶ < 2²⁰¹⁷

i don't now

mong thông cảm !

...........................

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

ta có :

\(\frac{1}{2^2}< \frac{1}{1\cdot2}\)

\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)

\(\frac{1}{4^2}< \frac{1}{3\cdot4}\)

...

\(\frac{1}{100^2}< \frac{1}{99\cdot100}\)

nên \(A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A< 1-\frac{1}{100}\)

\(\Rightarrow A< \frac{99}{100}< 1\)

\(\Rightarrow A< 1\left(đpcm\right)\)

nhiều qá lm sao nổi

a -35/50 = -7/10

b  510/2805 = 2/11

c  119/126

B2

-2/3= -8/12 , -1/4= -3/12

-8/12<-3/12 nên -2/3<-1/4

b 2/3  5/6

12/18 và 15/18

12/18<15/18

nên 14/21<60/72

bài 1 :

a) = -7/10

b) = 510/2805 = 2/11

c) = 17/18

`Answer:`

\(T=\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2016}{2^{2015}}+\frac{2017}{2^{2016}}\)

\(\Leftrightarrow2T=2+\frac{3}{2}+\frac{4}{2^2}+...+\frac{2016}{2^{2014}}+\frac{2017}{2^{2015}}\)

\(\Leftrightarrow2T-T=2+\left(\frac{3}{2}-\frac{2}{2}\right)+\left(\frac{4}{2^2}-\frac{4}{2^2}\right)+...+\left(\frac{2017}{2^{2015}}-\frac{2016}{2^{2015}}\right)-\frac{2017}{2^{2016}}\)

\(\Leftrightarrow2T-T=2+\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\frac{2017}{2^{2016}}\)

Ta đặt \(V=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)

\(\Rightarrow T=2+V-\frac{2017}{2^{2016}}\text{(*)}\)

\(\Leftrightarrow2V=1+\frac{1}{2}+...+\frac{1}{2^{2014}}\)

\(\Leftrightarrow2V-V=\left(1+\frac{1}{2}+...+\frac{1}{2^{2014}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)\)

\(\Leftrightarrow2V-V=1-\frac{1}{2^{2015}}\text{(**)}\)

Từ (*)(**)\(\Rightarrow T=2+\left(1-\frac{1}{2^{2015}}\right)-\frac{2017}{2^{2016}}\)

\(\Leftrightarrow T=3-\frac{1}{2^{2015}}-\frac{2017}{2^{2016}}\)

`=>T<3`

i giúp em vớiiiiii

\(M=\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+...+\dfrac{3}{1+2+...+2022}\)

\(=\dfrac{3}{\dfrac{2\left(2+1\right)}{2}}+\dfrac{3}{\dfrac{3\left(3+1\right)}{2}}+...+\dfrac{3}{\dfrac{2022\left(2022+1\right)}{2}}\)

\(=\dfrac{6}{2\left(2+1\right)}+\dfrac{6}{3\left(3+1\right)}+...+\dfrac{6}{2022\cdot2023}\)

\(=\dfrac{6}{2\cdot3}+\dfrac{6}{3\cdot4}+...+\dfrac{6}{2022\cdot2023}\)

\(=6\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2022\cdot2023}\right)\)

\(=6\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)

\(=6\cdot\left(\dfrac{1}{2}-\dfrac{1}{2023}\right)=6\cdot\dfrac{2021}{4046}=\dfrac{12126}{4046}< 3\)

mà \(3< \dfrac{10}{3}\)

nên \(M< \dfrac{10}{3}\)