Các câu hỏi tương tự
Cho A =\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)
B=\(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\)
C=\(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\)
Chứng minh A = B - 2C
Cho \(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(B=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}\)
\(C=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{48}+\frac{1}{50}\)
Chứng minh rằng A=B-2C
Chứng Minh Rằng:
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}\)
Chứng minh rằng : \(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
Chứng minh rằng : \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
Chứng minh rằng \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}\)
Chứng minh rằng l
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
Cho A= \(\frac{1}{1.2^2}+\frac{1}{2.3^2}+\frac{1}{3.4^2}+.....+\frac{1}{49.50^2}\)
B=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{50^2}\)
Chứng minh : A < \(\frac{1}{2}\)< B
chứng minh rằng \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)\(\frac{1}{50}\)
Chứng minh rằng:\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)