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C=1/15+1/35+1/63+..+1/2499
=1/3.5+1/5.7+1/7.9+...+1/49.51
=1/2(2/3.5+2/5.7+2/7.9+...+2/49.51)
=1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/49-1/51)
= 1/2.(1/3-1/51)
=1/2.16/51
=8/51
Trả lời
\(C=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(\Rightarrow C=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{...1}{49\cdot51}\)
\(\Rightarrow2C=2\left(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{49\cdot51}\right)\)
\(\Rightarrow2C=\frac{2}{1\cdot3}+\frac{2}{5\cdot7}+...+\frac{2}{49\cdot51}\)
\(\Rightarrow2C=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(\Rightarrow2C=1-\frac{1}{51}\)
\(\Rightarrow2C=\frac{50}{51}\)
\(\Rightarrow C=\frac{50}{51}:2\)
\(\Rightarrow C=\frac{25}{51}\)
Vậy \(C=\frac{25}{51}\)
Đặt \(A=1\frac{7}{15}-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}-\frac{1}{143}-\frac{1}{195}\)
\(\Rightarrow A=\frac{22}{15}-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\right)\)
Đặt \(B=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(\Rightarrow B=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\)
\(\Rightarrow2B=2\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\right)\)
\(\Rightarrow2B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}+\frac{2}{13\cdot15}\)
\(\Rightarrow2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(\Rightarrow2B=1-\frac{1}{15}\)
\(\Rightarrow2B=\frac{14}{15}\)
\(\Rightarrow B=\frac{14}{15}:2\Rightarrow B=\frac{7}{15}\)
\(\Rightarrow A=\frac{22}{15}-\frac{7}{15}\Rightarrow A=\frac{15}{15}=1\)
Ta có \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
Ta có \(B=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=\frac{1}{2}-\frac{1}{7}\)
\(=\frac{5}{14}\)
Ta có \(C=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{1}{6}-\frac{1}{22}\)
\(=\frac{4}{33}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(B=\frac{1}{2}-\frac{1}{7}\)
\(B=\frac{5}{14}\)
\(C=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(C=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(C=\frac{1}{6}-\frac{1}{22}=\frac{4}{33}\)
\(\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)=\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
=\(\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
=\(\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{13}\right)=\frac{1}{2}.\frac{8}{65}=\frac{4}{65}\)
Ta có :
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(=\)\(\frac{1}{2}\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{2499}\right)\) ( bước này hơi khó hiểu tí nhé )
\(=\)\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\) ( phân tích mẫu )
\(=\)\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\) ( áp dụng công thức thoi )
\(=\)\(\frac{1}{2}\left(1-\frac{1}{51}\right)\) ( loại bỏ nhưng phân số đối nhau )
\(=\)\(\frac{1}{2}.\frac{50}{51}\)
\(=\)\(\frac{25}{51}\)
Chúc bạn học tốt ~
ĐẶT \(A\)\(=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{49\cdot51}\)
\(2.A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{49\cdot51}\)
\(2.A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(2.A=1-\frac{1}{51}\)
\(2.A=\frac{50}{51}\)
\(\Rightarrow A=\frac{50}{51}\div2=\frac{25}{51}\)
Câu \(C=\left(...\right)\) thiếu đề
Ta có :
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(............\)
\(\frac{1}{10^2}< \frac{1}{9.10}\)
\(\Rightarrow\)\(D=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)
\(\Rightarrow\)\(D< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow\)\(D< 1-\frac{1}{10}< 1\)
\(\Rightarrow\)\(D< 1\) ( đpcm )
Vậy \(D< 1\)
Chúc bạn học tốt ~
\(C=\frac{1}{15}+\frac{1}{35}+....+\frac{1}{2499}\)
\(C=\frac{1}{3.5}+\frac{1}{5.7}+........+\frac{1}{49.51}\)
\(C=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{49.51}\right)\)
\(C=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+......+\frac{1}{49}-\frac{1}{51}\right)\)
\(C=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)\)
\(C=\frac{1}{2}.\frac{16}{51}\)
\(C=\frac{8}{51}\)
\(D=\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{10^2}\)
ta có :
\(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}\)
\(....................\)
\(\frac{1}{10^2}=\frac{1}{10.10}< \frac{1}{9.10}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{10^2}< \frac{1}{1.2}+\frac{1}{2.3}+........+\frac{1}{9.10}\)
\(\Rightarrow D< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+........+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow D< 1-\frac{1}{10}\)
\(\Rightarrow D< \frac{9}{10}\) ( 1 )
mà \(\frac{9}{10}< 1\) ( 2 )
từ ( 1 ) và ( 2 ) \(\Rightarrow D< 1\left(\text{đ}pcm\right)\)
\(C=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{49.51}\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)\)
\(C=\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{8}{51}\)