Các câu hỏi tương tự
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 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
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:\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+........+\frac{1}{99^2}+\frac{1}{100^2}< 1\frac{3}{4}\)
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) \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
Chứng minh rằng
a,\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}
Cho A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) .Chứng minh A<\(\frac{3}{4}\)