Bài 2 :

Ta có :

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{16}\right)+\left(\frac{1}{16}-\frac{1}{32}\right)\)

    \(=1-\frac{1}{32}\)

và \(B=\frac{2003}{2004}=1-\frac{1}{2004}\)

Vì \(\frac{1}{32}>\frac{1}{2004}\) nên A < B

A<B

sai đề

\(\frac{1}{1x2} +(\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9} +\frac{2}{9x11})\)

\(=\frac{1}{1x2} + (\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11})\)

\(=\frac{1}{1x2}+(\frac{1}{3}-\frac{1}{11})\)

\(=\frac{1}{1x2} +\frac{10}{33}\)

\(=\frac{1}{2} + \frac{10}{33} = \frac{33}{66}+\frac{20}{66}\)

\(=\frac{53}{66}\)

đúng mà

\(G=\frac{1}{3^0}+\frac{1}{3^1}+...+\frac{1}{3^{2005}}\)\(\Rightarrow3G=3+\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\)

\(\Rightarrow3G-G=2G=3-\frac{1}{3^{2005}}\)\(\Rightarrow G=\frac{3-\frac{1}{3^{2005}}}{2}\)

\(Y=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\)\(\Rightarrow2Y=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)

\(\Rightarrow2Y-Y=2-\frac{1}{2^{2012}}\) \(\Rightarrow Y=2-\frac{1}{2^{2012}}\)

Cảm ơn bạn nhiều lắm 

B=2/3x5 + 2/5x7 + 2/7x9 + ...+2/99x101

B= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 -1/9 + ... + 1/99 - 1/101

B= 1/3 - 1/101

B=98/303

( k mk nhé ! Cách làm câu a và b của mk đều đúng 100% đấy ! Dạng này mk học từ lâu rồi ! )

a, A = 1/2x3+ 1/ 3x4 + 1/4x5 + 1/5x6 + ... + 1/99x100

    A= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 -1/5 + 1/5 - 1/6 + ... + 1/99 -1/100

    A= 1/2 -1/100

    A= 49 / 100

nhờ bạn giải dùm mình câu c,d

a) \(\dfrac{2}{1\times4}+\dfrac{2}{4\times7}+\dfrac{2}{7\times10}+...+\dfrac{2}{97\times100}\)

\(=2.\left(\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+\dfrac{1}{7\times10}+...+\dfrac{1}{97\times100}\right)\)

\(=2.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

\(=2.\left(1-\dfrac{1}{100}\right)\)

\(=2.\dfrac{99}{100}\)

\(=\dfrac{99}{50}\)

_____

b) \(\dfrac{3}{1\times5}+\dfrac{3}{5\times9}+\dfrac{3}{9\times13}+...+\dfrac{3}{97\times101}\)

\(=3.\left(\dfrac{1}{1\times5}+\dfrac{1}{5\times9}+\dfrac{1}{9\times13}+...+\dfrac{1}{97\times101}\right)\)

\(=3.\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)

\(=3.\left(1-\dfrac{1}{101}\right)\)

\(=3.\dfrac{100}{101}\)

\(=\dfrac{300}{101}\)

dấu " . "  là dấu nhân nha bn, nãy viết nhanh quên lớp 

a=511/256

b=647/20

c=mình đang suy nghĩ,nhưng nếu bạn k cho mình thì bạn sẽ có câu trả lời

a. 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256

= 1 + ( 1 - 1/2) + ( 1/2 - 1/4) + ( 1/4 - 1/8) + ( 1/8 - 1/16) + ( 1/16 - 1/32) + (1/32 - 1/64) + ( 1/64 - 1/128) + (1/128 - 1/256)

= 1 + 1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128 + 1/128 - 1/256

= 2 - 1/256

= 511/256

Câu b bạn có viết sai đề không vậy?

b, 3/1x4 + 3/2x5 + 3/3x6 + 3/4x7 + 1/5x8

= 3/4 + 3/10 + 3/18 + 3/28 + 1/40

= 1133/840

c,2/1x3 + 2/3x5 + 2/5x7+..+ 2/2001x2003 + 2/2003x2005

= ( 1 - 1/3) + ( 1/3 - 1/5) + ( 1/5 - 1/7) +...+ ( 1/2001 - 1/2003) + (1/2003 - 1/2005)

= 1 - 1/3 +  1/3 - 1/5 + 1/5 - 1/7 +...+  1/2001 - 1/2003 + 1/2003 - 1/2005

= 1 - 1/2005

= 2004/2005

\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{13\times15}+\dfrac{2}{15\times17}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{17}\)

\(=1-\dfrac{1}{17}\)

\(=\dfrac{16}{17}\)

\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{15\cdot17}\)

\(=2-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{17}\)

\(=2-\dfrac{1}{17}\)

\(=\dfrac{33}{17}\)