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

\(C=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{49.51}\)

\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)\)

\(C=\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)

\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{8}{51}\)

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 ) \(\Rightarrow D< 1\left(\text{đ}pcm\right)\)

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 \(C=\frac{25}{51}\)

\(=\frac{16}{51}\)

a)\(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}=\frac{5}{3}\cdot\left(\frac{3}{1.4}+\frac{4}{4.7}+...+\frac{3}{100.103}\right)\)

\(=\frac{5}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}\cdot\left(1-\frac{1}{103}\right)=\frac{5}{3}\cdot\frac{102}{103}=\frac{170}{103}\)b)\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

\(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{1}{2}\cdot\frac{16}{51}=\frac{8}{51}\)

Câu a) bạn Ác Mộng làm rồi nên mình làm b) nha

b)Gọi A = \(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)

\(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

\(2A=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\right)\)

\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)

\(2A=\frac{1}{3}-\frac{1}{51}\)

\(2A=\frac{16}{51}\)

\(A=\frac{16}{51}:2\)

\(A=\frac{8}{51}\)

\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}=\frac{1}{2}.(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51})\)

\(=\frac{1}{3}-\frac{1}{51}=\frac{16}{51}\)

\(C=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)

\(C=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

\(C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)

\(C=\frac{1}{3}-\frac{1}{51}\)

\(C=\frac{16}{51}\)

\(A=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{16}{15}\right)...\left(1+\frac{2500}{2499}\right)\)

\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}...\frac{2500}{2499}\)

\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{50.50}{49.51}\)

\(=2.\frac{2}{3}.\frac{3}{2}.\frac{3}{4}.\frac{4}{3}.\frac{4}{5}...\frac{50}{49}.\frac{50}{51}\)

\(=\left(2.\frac{3}{2}.\frac{4}{3}...\frac{50}{49}\right)\left(\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{50}{51}\right)\)

\(=\frac{2.3.4...50}{2.3.4...49}.\frac{2.3.4...50}{3.4.5...51}\)

\(=50.\frac{2}{51}=\frac{100}{51}\)

a) \(\frac{1}{n}-\frac{1}{n+a}=\frac{\left(n+a\right)-n}{n\left(n+a\right)}=\frac{a}{a\left(n+a\right)}\) (đpcm)

b) \(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

\(B=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}.\left(1-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)

\(C=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}=\frac{1}{3}-\frac{1}{51}=\frac{16}{51}\)