A=1/2^2+1/3^2+...+1/100^2
=>A<1-1/2+1/2-1/3+...+1/99-1/100
=>A<1-1/100<1
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25 tháng 4 2023 lúc 9:53
Chứng tỏ rằng:
a)\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{3}{4}\)
b)\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
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 , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
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 \(F< 1\dfrac{3}{4}\)
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
\(\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}\)
25 tháng 2 2022 lúc 9:14
Tính:
S = \(\dfrac{1}{2}\) - \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) - \(\dfrac{1}{2^4}\) + ... + \(\dfrac{1}{2^{99}}\) - \(\dfrac{1}{2^{100}}\)
A= \(\dfrac{1}{2}\)-\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)-\(\dfrac{1}{2^4}\)+...+\(\dfrac{1}{2^{99}}\)-\(\dfrac{1}{2^{100}}\)
22 tháng 3 2023 lúc 11:39
\(A=\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}\)
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)
A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{98^2}+\dfrac{1}{99^2}+\dfrac{1}{100^2}< 1\)