A = 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^n < 1

A = 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^n

2A = 1 + 1/2 + 1/2^2+ ..+ 1/2^n-1

2A - A = 1 + 1/2 + 1/2^2+ ..+ 1/2^n-1 - (1/2 + 1/2^2 + 1/2^3 + ... + 1/2^n)

A = (1 - 1/2^n) + (1/2 - 1/2) + ..+ (1/2^n-1 -1/2^n-1)

A = 1 - 1/2^n

A < 1 (đpcm)

Câu b:

B = 1/3 + 1/3^2 + 1/3^3 + ...+ 1/3^n < 1/2

3B = 1 + 1/3 + 1/3^2 + ..+ 1/3^n - 1

3B - B = 1 + 1/3 + 1/3^2 + ..+ 1/3^n - 1 - (1/3 + 1/3^2 + 1/3^3 + ...+ 1/3^n)

2B = 1 + 1/3 + 1/3^2 + ..+ 1/3^n - 1 - 1/3 - 1/3^2 - 1/3^3 -..- 1/3^n-1 - 1/3^n

2B = (1 - 1/3^n) + (1/3 - 1/3) +..+(1/3^n-1-1/3^n-1)

2B = 1 - 1/3^n

B = 1/2 - 1/2.3^n < 1/2 (đpcm)

a: \(A=\dfrac{1-\dfrac{1}{\sqrt{49}}+\dfrac{1}{49}-\dfrac{1}{\left(7\sqrt{7}\right)^2}}{\dfrac{\sqrt{64}}{2}-\dfrac{4}{7}+\left(\dfrac{2}{7}\right)^2-\dfrac{4}{343}}\)

\(=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}\)

\(=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4\left(1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}\right)}=\dfrac{1}{4}\)

b: \(M=1-\dfrac{5}{\sqrt{196}}-\dfrac{5}{\left(2\sqrt{21}\right)^2}-\dfrac{\sqrt{25}}{204}-\dfrac{\left(\sqrt{5}\right)^2}{374}\)

\(=1-\dfrac{5}{14}-\dfrac{5}{84}-\dfrac{5}{204}-\dfrac{5}{374}\)

\(=1-5\left(\dfrac{1}{14}+\dfrac{1}{84}+\dfrac{1}{204}+\dfrac{1}{374}\right)\)

\(=1-5\left(\dfrac{1}{2\cdot7}+\dfrac{1}{7\cdot12}+\dfrac{1}{12\cdot17}+\dfrac{1}{17\cdot22}\right)\)

\(=1-\left(\dfrac{5}{2\cdot7}+\dfrac{5}{7\cdot12}+\dfrac{5}{12\cdot17}+\dfrac{5}{17\cdot22}\right)\)

\(=1-\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{22}\right)\)

\(=1-\left(\dfrac{1}{2}-\dfrac{1}{22}\right)\)

\(=1-\dfrac{11-1}{22}=1-\dfrac{10}{22}=\dfrac{12}{22}=\dfrac{6}{11}\)

a) A =  1 - 2 + 3 - 4 + .....+ 2013 - 2014 + 2015

=  (1 - 2) + (3 - 4) + .....+ (2013 - 2014) + 2015

=(-1) + (-1) + ...+ (-1) + 2015

      1007 số  -1

=(-1) . 1007 +2015   

= -1007 + 2015

=1008

b)  1 . 2 + 2 . 3 + 3 . 4 +....99 . 100

đặt S= 1 . 2 + 2 . 3 + 3 . 4 +....99 . 100

 => 3S = 1.2.3 +2.3.(4-1) +...+99.100.(101-98)

 = 1.2.3 +2.3.4-1.2.3+...+99.100.101-98.99.100

= 99.100.101

= 999900

=> S= 333300

Vậy 1 . 2 + 2 . 3 + 3 . 4 +....99 . 100 = 333300

a,A=(1-2)+(3-4)....(2013-2014)+2015

A= -1 + -1.....-1+2015

A= (2015-1):1+1

A=2015 

A=(2015 x -1) x -1

A=2015

A=2015 + 2015 

A=4030

b, 1/1.2 +1/2.3 ...1/99.100

1/1-1/2+1/2-1/3 ....1/99-1/100

1/1-1/100

99/100

a: Ta có công thức tổng quát:

\(1-\frac{1}{1+2+\cdots+n}\)

\(=1-\frac{1}{\frac{n\left(n+1\right)}{2}}=1-\frac{2}{n\left(n+1\right)}\)

\(=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)

Ta có: \(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\cdot\ldots\cdot\left(1-\frac{1}{1+2+\cdots+2022}\right)\)

\(=\frac{\left(2+2\right)\left(2-1\right)}{2\left(2+1\right)}\cdot\frac{\left(3+2\right)\left(3-1\right)}{3\left(3+1\right)}\cdot\ldots\cdot\frac{\left(2022+2\right)\left(2022-1\right)}{2022\left(2022+1\right)}\)

\(=\frac{4\cdot5\cdot\ldots\cdot2024}{3\cdot4\cdot\ldots\cdot2023}\cdot\frac{1\cdot2\cdot\ldots\cdot2021}{2\cdot3\cdot\ldots\cdot2022}=\frac{2024}{3}\cdot\frac{1}{2022}=\frac{1012}{1011\cdot3}=\frac{1012}{3033}\)

b:Sửa đề: \(B=1+\frac12\left(1+2\right)+\frac13\left(1+2+3\right)+\cdots+\frac{1}{100}\left(1+2+\cdots+100\right)\)

\(=1+\frac12\cdot\frac{2\cdot3}{2}+\frac13\cdot\frac{3\cdot4}{2}+\cdots+\frac{1}{100}\cdot\frac{100\cdot101}{2}\)

\(=1+\frac32+\frac42+\cdots+\frac{101}{2}=\frac12\left(2+3+4+\cdots+101\right)\)

\(=\frac12\left(101-2+1\right)\cdot\frac{101+2}{2}=\frac12\cdot100\cdot\frac{101+2}{2}=103\cdot25=2575\)

khó nhìn lắm bn ak

sao pn ko cho 

\(\frac{11}{125}-\frac{17}{18}-\frac{5}{8}+\frac{4}{9}+\frac{17}{14}.\)

thì có phải dễ nhìn hơn ko

a, \(A=\frac{11}{125}-\frac{17}{18}-\frac{5}{7}+\frac{4}{9}+\frac{17}{14}\)

\(=\frac{11}{125}+\left(\frac{-17}{18}+\frac{4}{9}\right)+\left(\frac{-5}{7}+\frac{17}{14}\right)\)

\(=\frac{11}{125}+\frac{-1}{2}+\frac{1}{2}\)

\(=\frac{11}{125}\)

b, \(B=1-\frac{1}{2}+2-\frac{2}{3}+3-\frac{3}{4}+4-\frac{1}{4}-3-\frac{1}{3}-2-\frac{1}{2}-1\)

\(=\left(1+2+3+4-3-2-1\right)-\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\right)\)

\(=4-3=1\)

c, \(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

\(=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}=\frac{-49}{50}\)

Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)

\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)

\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)

\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)

\(=2017\)

Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)

\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)

\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)

\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)

\(=2017\)