Các câu hỏi tương tự
18 tháng 3 2023 lúc 19:38
chứng minh rằng
a , \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{512}-\dfrac{1}{1024}\) < \(\dfrac{1}{3}\)
b , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
22 tháng 3 2023 lúc 11:39
\(A=\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}\)
A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{98^2}+\dfrac{1}{99^2}+\dfrac{1}{100^2}< 1\)
9 tháng 5 2021 lúc 20:08
\(\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}\)
\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\)
94-\(\dfrac{1}{7}-\dfrac{2}{8}-\dfrac{3}{9}-...-\dfrac{94}{100}\)
\(\dfrac{1}{35}+\dfrac{1}{40}+\dfrac{1}{45}+...+\dfrac{1}{500}\)
giúp mik nha mik cần gấp
\(\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}\)
18 tháng 2 2022 lúc 18:27
Cho F = \(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}=\dfrac{1}{4^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}\). Chứng tỏ \(F< 1\dfrac{3}{4}\)
25 tháng 2 2022 lúc 9:14
Tính:
S = \(\dfrac{1}{2}\) - \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) - \(\dfrac{1}{2^4}\) + ... + \(\dfrac{1}{2^{99}}\) - \(\dfrac{1}{2^{100}}\)
11 tháng 4 2021 lúc 15:06
text{Bài 4. Chứng tỏ rằng:}a) dfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+...+dfrac{1}{30^2} 1b) dfrac{1}{10}+dfrac{1}{11}+dfrac{1}{12}+...+dfrac{1}{99}+dfrac{1}{100}1c) dfrac{1}{5}+dfrac{1}{6}+dfrac{1}{7}+...+dfrac{1}{17} 2d) dfrac{1}{1.2}+dfrac{1}{2.3}+dfrac{1}{3.4}+...+dfrac{1}{29.30} 1
Đọc tiếp
\(\text{Bài 4. Chứng tỏ rằng:}\)
\(a\)) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}< 1\)
\(b\)) \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}>1\)
\(c\)) \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)
\(d\)) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}< 1\)
B(1-dfrac{1}{2^2})(1-dfrac{1}{3^2})(1-dfrac{1}{4^2})(1-dfrac{1}{5^2})...(1-dfrac{1}{99^2})(1-dfrac{1}{100^2})C(dfrac{1}{4}-1)(dfrac{1}{9}-1)(dfrac{1}{16}-1)...(dfrac{1}{999}-1)
Đọc tiếp
B=(1-\(\dfrac{1}{2^2}\))(1-\(\dfrac{1}{3^2}\))(1-\(\dfrac{1}{4^2}\))(1-\(\dfrac{1}{5^2}\))...(1-\(\dfrac{1}{99^2}\))(1-\(\dfrac{1}{100^2}\))
C=(\(\dfrac{1}{4}\)-1)(\(\dfrac{1}{9}\)-1)(\(\dfrac{1}{16}\)-1)...(\(\dfrac{1}{999}\)-1)