1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 3/4 (đpcm)
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 3/4 (đpcm)
Chứng tỏ rằng:
a)\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{3}{4}\)
b)\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
Chứng tỏ rằng: A= 2+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}^{ }\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{100^2}\)<3
cho A=\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+\(\dfrac{1}{2^4}\)+.....+\(\dfrac{1}{2^{2020}}\)+\(\dfrac{1}{2^{2021}}\). Chứng tỏ rằng A<\(\dfrac{1}{2}\)
Giúp vs ạ cần gấp
Cho \(S=\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+...+\dfrac{99}{5^{100}}\). Chứng tỏ rằng S<\(\dfrac{1}{16}\)
Cho S=\(\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+...+\dfrac{99}{5^{100}}\) . Chứng tỏ rằng \(S< \dfrac{1}{16}\)
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 \(F< 1\dfrac{3}{4}\)
Chứng minh rằng: \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^3}\)+\(\dfrac{1}{4^2}+....+\dfrac{1}{100^2}< \dfrac{3}{4}\)
Giúp mình với.Mình cần gấp ạ
\(C=\)\(\dfrac{-5}{7}+\dfrac{3}{4}+\dfrac{-1}{5}+\dfrac{-2}{7}+\dfrac{1}{4}\)
GIẢI CHI TIẾT HỘ MÌNH VỚI!!!MÌNH CẦN GẤP!!!!! MÌNH CẢM ƠN NHIỀU Ạ!!!!
\(\text{Bài 4. Chứng tỏ rằng:}\)
\(a\)) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}< 1\)
\(b\)) \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}>1\)
\(c\)) \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)
\(d\)) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}< 1\)