\(a,\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

\(=1-\frac{1}{7}\)

\(=\frac{6}{7}\)

\(b,\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)

Ta có :

\(\frac{1}{2}=1-\frac{1}{2}\)

\(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)

\(\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\)

\(\frac{1}{16}=\frac{1}{8}-\frac{1}{16}\)

\(\frac{1}{32}=\frac{1}{16}-\frac{1}{32}\)

Thay vào ta có :

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

\(=1-\frac{1}{32}\)

\(=\frac{31}{32}\)

\(c,\)\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)

Ta có :

\(\frac{1}{2}=1-\frac{1}{2}\)

\(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)

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

\(\frac{1}{256}=\frac{1}{128}-\frac{1}{256}\)

Thay vào ta có :

\(=\)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{128}-\frac{1}{256}\)

\(=1-\frac{1}{256}\)

\(=\frac{255}{256}\)

a: \(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^7\)

=>\(2\cdot A=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^6\)

=>\(2A-A=1-\left(\dfrac{1}{2}\right)^7=1-\dfrac{1}{128}=\dfrac{127}{128}\)

=>\(A=\dfrac{127}{128}\)

b: \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{10\cdot11}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\)

\(=1-\dfrac{1}{11}=\dfrac{10}{11}\)

C=1/2+1/4+1/8+1/16+1/32+1/64+1/128

C=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

C=1-1/128

C=127/128

D=1/2+1/6+1/12+1/20+1/30+1/42

D=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7

D=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7

D=1/1-1/7

D=6/7

\(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)

\(2B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)

\(2B-B=1-\frac{1}{64}\)

\(B=\frac{63}{64}\)

A=1/1*2+1/2*3+1/3*4+1/4*5+1/5*6+1/6*7

=1-1/2+1/2-1/3+1/3-1/4+1/4

=1/5+1/5-1/6+1/6-1/7

=1-1/7

=6/7

cauB tương tự bạn tự làm nhé 

CHÚC BẠN HỌC TỐT

\(B=\frac{6}{1\cdot3}+\frac{6}{3\cdot5}+\cdot\cdot\cdot+\frac{6}{97\cdot99}\)

\(\Rightarrow B=3\cdot\left(\frac{2}{1\cdot3}+\cdot\cdot\cdot+\frac{2}{97\cdot99}\right)\)

\(\Rightarrow B=3\cdot\left(1-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{97}-\frac{1}{99}\right)\)

\(\Rightarrow B=3\cdot\left(1-\frac{1}{99}\right)\)

\(\Rightarrow B=3\cdot\frac{98}{99}\)

\(\Rightarrow B=\frac{98}{33}\)

\(A=\frac{1}{2}+\frac{1}{6}+\cdot\cdot\cdot+\frac{1}{42}\)

\(\Rightarrow A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{6\cdot7}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{6}-\frac{1}{7}\)

\(\Rightarrow A=1-\frac{1}{7}\)

\(\Rightarrow A=\frac{6}{7}\)

a.A= 1/2 + 1/4+ 1/8+ 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512 
A = 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 - 1/256 - 1/512 
A = 1 - 1/512 
A = 511/512 

b. 1/2 + 1/6 + 1/12 + … + 1/110
= 1/1.2 + 1/2.3 + 1/3.4 + … + 1/10.11. (dấu . thay dấu x).
= 1/1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 +…+ 1/10 – 1/11
= 1/1 – 1/11
= 10/11

Chúc bạn học giỏi nha!

a ) Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{512}\)

\(\Rightarrow A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)

Nhân 2 vào hai vế của biểu thức A , ta được :

\(2A=2.\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)

\(\Rightarrow2A=1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^8}\)

Lấy biểu thức 2A - A , ta được :

\(2A-A=\left(1+\frac{1}{2^1}+\frac{1}{2^2}+....+\frac{1}{2^8}\right)-\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)

\(\Rightarrow A=1-\frac{1}{2^9}\Rightarrow A=\frac{512}{512}-\frac{1}{512}=\frac{511}{512}\)

Vậy \(A=\frac{511}{512}\)

b ) Đặt \(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{90}+\frac{1}{110}\)

\(\Rightarrow B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}+\frac{1}{10.11}\)

\(\Rightarrow B=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

\(B=1-\frac{1}{11}=\frac{11}{11}-\frac{1}{11}=\frac{10}{11}\)

Vậy \(B=\frac{10}{11}\)

a) \(\dfrac{3}{8}+\dfrac{7}{12}+\dfrac{10}{16}+\dfrac{10}{24}\)

\(=\dfrac{3}{8}+\dfrac{7}{12}+\dfrac{5}{8}+\dfrac{5}{12}\)

\(=\left(\dfrac{3}{8}+\dfrac{5}{8}\right)+\left(\dfrac{7}{12}+\dfrac{5}{12}\right)\)

\(=1+1\)

\(=2\)

b) \(\dfrac{4}{6}+\dfrac{7}{13}+\dfrac{17}{9}+\dfrac{19}{13}+\dfrac{1}{9}+\dfrac{14}{6}\)

\(=\dfrac{2}{3}+\dfrac{7}{13}+\dfrac{17}{9}+\dfrac{19}{13}+\dfrac{1}{9}+\dfrac{7}{3}\)

\(=\left(\dfrac{2}{3}+\dfrac{7}{3}\right)+\left(\dfrac{7}{13}+\dfrac{19}{13}\right)+\left(\dfrac{17}{9}+\dfrac{1}{9}\right)\)

\(=3+2+2\)

\(=7\)

c) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)

\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)

\(=1-\dfrac{1}{7}\)

\(=\dfrac{6}{7}\)

xem lại câu c)

a) 3/8  b) 31/32

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{56}=\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}...\frac{1}{7x8}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)\(-\frac{1}{8}=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)

b,