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

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

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

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

Ta có:\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\)\(\frac{1}{64}\)

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

     \(=1-\frac{1}{64}\)\(=\frac{63}{64}\)

Cách 1:

B=1/2+1/4+1/8+1/16+1/32+1/64

B=1-1/2 + 1/2-1/4 + 1/4-1/8 +1/8-1/16 + 1/16-1/32 + 1/32-1/64

B=1-1/64

B=63/64

Cách 2:

B=1/2+1/4+1/8+1/16+1/32+1/64

B=1/2¹+1/2²+1/2³+1/2⁴+1/2⁵+1/2⁶

2B=1+1/2¹+1/2^2+1/2^3+1/2^4+1/2^5

2B-B=1-1/2^6

B=1-1/64 

B=63/64

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

=\(\frac{16+8+4+2+1}{32}=\frac{31}{32}\)

1/2+1/4+1/6+1/8+1/16):x=3/1*2+3/2*3+3/3*4+...+3/15*16

\(\frac{53}{48}:x=3(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{15.16})\)

\(\frac{53}{48}:x=3(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{15}-\frac{1}{16})\)

\(\frac{53}{48}:x=3(1-\frac{1}{16})\)

\(\frac{53}{48}:x=\frac{45}{16}\)

\(x=\frac{53}{48}:\frac{45}{16}\)

\(x=\frac{53}{135}\)

Chúc bạn học tốt

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

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

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

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

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

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

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 :

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

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

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

b)Ta có:\(A=1+\frac{1}{2.\left(1+2\right)}+\frac{1}{3.\left(1+2+3\right)}+...+\frac{1}{16.\left(1+2+3+...+16\right)}\)

                 \(=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{16}.\left(1+2+3+...+16\right)\)

                 \(=1+\frac{1}{2}.3+\frac{1}{3}.6+...+\frac{1}{16}.136\)

                 \(=1+1,5+2+...+8,5\)

                 \(=\frac{\left(8,5+1\right).\left[\left(8,5-1\right):0,5+1\right]}{2}=76\)

B=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}<\)                                                                               

 B=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

B=\(1-\frac{1}{8}=\frac{8}{8}-\frac{7}{8}=\frac{1}{8}<2\)

Vậy 1/8<2 hay 1/8<16/8

A = 1/2^2 + 1/3^2 +.. + 1/8^2 < 1/1.2 + 1/2.3  +... + 1/7.8 = 1 - 1/2 + 1/2 -1/3 +...  + 1/7 - 1/8

=  1 - 1/8 < 1 

\(\Rightarrowđpcm\)

\(tíchnhaminhftchlaij\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{32}-\frac{1}{64}\)

\(=\frac{1}{1}-\frac{1}{64}=\frac{63}{64}\)

a) \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{64}\)

\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{32}\)

\(\Rightarrow2A-A=A=1-\frac{1}{64}=\frac{63}{64}\)