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\(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{28}\)
=> \(\frac{16}{64}+\frac{8}{64}+\frac{4}{64}+\frac{2}{64}+\frac{1}{64}+\frac{1}{28}\)
=> \(\frac{31}{64}+\frac{1}{28}=>\frac{217}{448}+\frac{16}{448}=\frac{233}{448}\)
a) Ta có : \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{4^2}< \frac{1}{3\cdot4}\)
. . .
\(\frac{1}{100^2}< \frac{1}{99\cdot100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2^2}\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\right)\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{4}\left(1+1-\frac{1}{50}\right)\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{4}\cdot\frac{99}{50}=\frac{99}{200}< \frac{100}{200}=\frac{1}{2}\left(đpcm\right)\)
b) Ta có :
\(B=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}>48\)
\(\Rightarrow1-\frac{1}{4}+1-\frac{1}{9}+...+1-\frac{1}{2500}>48\)
\(\Rightarrow49-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< 49\)
Lại có : \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)
. . .
\(\frac{1}{50^2}< \frac{1}{49\cdot50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow\frac{1}{2^2}+...+\frac{1}{50^2}< \frac{49}{50}< 1\)
\(\Rightarrow-\left(\frac{1}{2^2}+...=\frac{1}{50^2}\right)>1\)
\(\Rightarrow49-\left(\frac{1}{2^2}+...+\frac{1}{50^2}\right)>49-1=48\)
hay \(\frac{3}{4}+\frac{8}{9}+...+\frac{2499}{2500}>48\left(đpcm\right)\)
A=1/8+1/24+1/48+1/80+1/120+1/168+1/224=>2A=2/8+2/24+2/48+2/80+2/120+2/168+2/224
2A=2/2*4+2/4*6+2/6*8+2/8*10+2/10*12+2/12*14+2/14*16
2A=1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10+1/10-1/12+1/12-1/14+1/14-1/16
2A=1/2-1/16
2A=7/16
A=7/16:2
A=7/32
\(A=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
TA có :\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{50^2}< \frac{1}{49.50}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+1=2\)
\(A=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< \frac{1}{2^2}.2=\frac{1}{2}\left(đpcm\right)\)
\(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{24}+\frac{1}{48}+\frac{1}{28}\)
=> \(\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)+\left(\frac{1}{24}+\frac{1}{48}\right)+\frac{1}{28}\)
=> \(\left(\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right)+\left(\frac{2}{48}+\frac{1}{48}\right)+\frac{1}{28}\)
=> \(\frac{7}{16}+\frac{3}{48}+\frac{1}{28}\)
=> \(\frac{1}{2}+\frac{1}{28}=>\frac{14}{28}+\frac{1}{28}=\frac{15}{28}\)
đúng 100%