Các câu hỏi tương tự
cho \(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\)
chung to rang B >1
cho B = .\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+.....+\frac{1}{19}\) Hay chung to B< 1
Cho B=1/4+1/5+1/6+...+1/19.Hay chung to rang B>1
Cho A = \(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{2012^2}+\frac{1}{2013^2}\)
Hay Chung to rang A < 1
NHANH LEN NHA CAC BAN!
Chung to rang
b) B = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}< \frac{1}{2}\)
chung minh rang B=\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)\(\frac{1}{7^2}\)+\(\frac{1}{4^2}\)+\(\frac{1}{5^2}\)+\(\frac{1}{6^2}\)+\(\frac{1}{8^2}\)<1
Cho S = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7\cdot10}+....+\frac{3}{40\cdot43}+\frac{3}{43\cdot46}\) . Hay chung to rang S < 1
Bài 5: Cho B = \(\frac{1}{4} +\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\). Hãy chứng tỏ B > 1.
Cho B= \(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\). Chứng minh B>1