Các câu hỏi tương tự
Chứng tỏ rằng:
a) \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}<2\)
b) \(B=\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{39}+\frac{1}{40}.\) Chứng tỏ \(\frac{1}{2}\)< B < 1
c) \(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}<\frac{1}{100}\)
Chứng minh:
c.\(\frac{11}{15}< \frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{59}+\frac{1}{60}< \frac{3}{2}\)
b.\(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+\frac{1}{41}+\frac{1}{61}+\frac{1}{85}+\frac{1}{113}< \frac{1}{2}\)
a.\(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}< \frac{1}{2}\)
Chứng minh rằng:
a/\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\)
b/\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 1\frac{3}{4}\)
c/\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
Chứng minh rằng :
\(100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+...+\frac{99}{100}\)
Chứng minh rằng:
a,\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b,\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
giúp minh với
$frac{1}{2^2}+frac{1}{3^2}+frac{1}{4^2}+...+frac{1}{100^2}$122 +132 +142 +...+11002 Chứng minh A2 $frac{1}{2^2}+frac{1}{3^2}+frac{1}{4^2}+...+frac{1}{100^2}$122 +132 +142 +...+11002 Chứng minh A2 $frac{1}{2^2}+frac{1}{3^2}+frac{1}{4^2}+...+frac{1}{100^2}$122 +132 +142 +...+11002 Chứng minh A2$frac{1}{2^2}+frac{1}{3^2}+frac{1}{4^2}+...+frac{1}{100^2}$122 +132 +142 +...+11002 Chứng minh A2
Đọc tiếp
$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}$122 +132 +142 +...+11002
Chứng minh A<2
$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}$122 +132 +142 +...+11002
Chứng minh A<2
$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}$122 +132 +142 +...+11002
Chứng minh A<2
$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}$122 +132 +142 +...+11002
Chứng minh A<2
Chứng minh
100-(1-\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\))= \(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
Chứng minh:\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+........+\frac{1}{99^2}+\frac{1}{100^2}< 1\frac{3}{4}\)
Bài 1 : Chứng minh
a) \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 2\)
b) \(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 6\)
c) \(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.\frac{7}{8}...\frac{9999}{10000}< \frac{1}{100}\)