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lúc đầu ý bn là 5/1.3 đúng k, mk chỉnh lại như thế cho tiện nhé
a) \(\frac{5}{1\times3}+\frac{5}{3\times5}+\frac{5}{5\times7}+...+\frac{5}{99\times101}\)
\(=\frac{5}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}\left(1-\frac{1}{101}\right)\)
\(=\frac{5}{2}\times\frac{100}{101}=\frac{250}{101}\)
b) \(\frac{3^2}{8\times11}+\frac{3^2}{11\times14}+\frac{3^2}{14\times17}+...+\frac{3^2}{197\times200}\)
\(=\frac{9}{8\times11}+\frac{9}{11\times14}+\frac{9}{14\times17}+...+\frac{9}{197\times200}\)
\(=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(=3\left(\frac{1}{8}-\frac{1}{200}\right)\)
\(=3\times\frac{3}{25}=\frac{9}{25}\)
Ta có \(\frac{3^2}{8.11}+\frac{3^2}{11.14}+...+\frac{3^2}{197.200}\)
\(\Rightarrow3^2.\left(\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{197.200}\right)\)
\(\Rightarrow9.\frac{1}{3}.\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(\Rightarrow3.\left(1-\frac{1}{200}\right)\)
\(\Rightarrow3.\frac{199}{200}=\frac{597}{200}\)
#)Giải :
\(A=\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times\frac{24}{25}\times...\times\frac{2499}{2500}\)
\(A=\frac{1.3}{2.2}\times\frac{2.4}{3.3}\times\frac{3.5}{4.4}\times\frac{4.6}{5.5}\times...\times\frac{49.51}{50.50}\)
\(A=\frac{1\times3\times2\times4\times3\times5\times...\times49\times51}{2\times2\times3\times3\times4\times4\times...\times50\times50}\)
\(A=\frac{1\times51}{2\times50}\)
\(A=\frac{51}{100}\)
\(A=\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times\frac{24}{25}\times...\times\frac{2499}{2500}\)
\(=\frac{1\times3}{2\times2}\times\frac{2\times4}{3\times3}\times\frac{3\times5}{4\times4}\times\frac{6\times4}{5\times5}\times...\times\frac{49.51}{50\times50}\)
\(=\frac{1}{2}\times\frac{51}{50}\)
\(=\frac{51}{100}\)
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. Đặt A= biểu thức đã cho
=>\(\frac{A}{3}=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
=>\(\frac{A}{3}.2=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
=>\(\frac{2A}{3}-\frac{A}{3}=2-\frac{1}{2^9}\)
=>\(A=\frac{3\left(2^{10}-1\right)}{2^9}\)
B. Đặt B=biểu thức đã cho
\(\Rightarrow\frac{B}{2}=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2015.2017}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)
\(=\frac{1}{3}-\frac{1}{2017}=\frac{2014}{6051}\)
\(\Rightarrow B=\frac{4028}{6051}\)
\(A=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
TA có :\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{50^2}< \frac{1}{49.50}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+1=2\)
\(A=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< \frac{1}{2^2}.2=\frac{1}{2}\left(đpcm\right)\)
\(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)
\(=\frac{1}{3}-\frac{1}{21}\)
\(=\frac{7}{21}-\frac{1}{21}=\frac{6}{21}\)
\(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)
\(A=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+\left(\frac{1}{9}-\frac{1}{9}\right)+...+\left(\frac{1}{19}-\frac{1}{19}\right)-\frac{1}{21}\)
\(A=\frac{1}{3}-\frac{1}{21}\)
\(A=\frac{2}{7}\)