VIẾT SAI ĐỀ BÀI NHÉ
50<A<100
Các câu hỏi tương tự
chứng minh rằng \(50< ,1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2^{100}-1}< 100\)
1.chứng minh rằng : \(\frac{1}{2}!+\frac{2}{3}!+\frac{3}{4}!+...+\frac{99}{100}!< 1\)
2. Chứng minh rằng :\(\frac{1.2-1}{2}+\frac{2.3-1}{3}+\frac{3.4-1}{4}+...+\frac{99.100-1}{100}< 2\)
Chứng minh rằng:
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\) \(B=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}< \frac{3}{4}\)
chứng minh rằng \(50< \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2^{100}-1}< 100\)
việt anh hoàng hay ai giải cho tui với nha
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 \(\frac{1}{1}-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-............-\frac{1}{100^2}< \frac{1}{100}\)
chứng minh rằng : \(\frac{1}{7^2}-\frac{1}{7^4}+\frac{1}{7^6}-...+\frac{1}{7^{98}}-\frac{1}{7^{100}}< \frac{1}{50}\)
Chứng minh rằng:a. frac{1}{3^2}+frac{2}{3^3}+frac{3}{3^4}+frac{4}{3^5}+...+frac{99}{3^{100}}+frac{100}{3^{101}} frac{1}{4}b.frac{1}{2}-frac{1}{4}+frac{1}{8}-frac{1}{16}+frac{1}{32}-frac{1}{64} frac{1}{3}c.frac{1}{3}-frac{2}{3^2}+frac{3}{3^3}-frac{4}{3^4}+...+frac{99}{3^{99}}-frac{100}{3^{100}} frac{1}{16}d. frac{1}{5^2}-frac{2}{5^3}+frac{3}{5^4}-frac{4}{5^5}+...+frac{99}{5^{100}}-frac{100}{5^{101}} frac{1}{36}
Đọc tiếp
Chứng minh rằng:
a. \(\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\frac{4}{3^5}+...+\frac{99}{3^{100}}+\frac{100}{3^{101}}< \frac{1}{4}\)
b.\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
c.\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{1}{16}\)
d. \(\frac{1}{5^2}-\frac{2}{5^3}+\frac{3}{5^4}-\frac{4}{5^5}+...+\frac{99}{5^{100}}-\frac{100}{5^{101}}< \frac{1}{36}\)
chứng minh rằng:
a) A= \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)<1
b)B=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{4^4}+...+\frac{100}{3^{100}}< \frac{3}{4}\)