Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2}...\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Rightarrow A< \dfrac{1}{1.2}+...+\dfrac{1}{99.100}\)
\(\Rightarrow A< 1-\dfrac{1}{2}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow A< 1-\dfrac{1}{100}\)
\(\Rightarrow A< \dfrac{99}{100}\) Vì \(\dfrac{99}{100}< 1\Rightarrow A< 1\)
Chứng Minh Rằng : A= \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{100^2}\) <\(\dfrac{3}{4}\)
chứng minh rằng : \(\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}\)
Bài 7:
Chứng minh rằng: \(\dfrac{3}{10}< \dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}.\)
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 , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
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
Chứng tỏ rằng:
a)\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{3}{4}\)
b)\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
Cho: A= 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{2^{100}-1}\)
Chứng minh rằng: 50 < A < 100
Giúp mình với!
Chứng minh rằng \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)
\(S=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\) Chứng minh rằng \(S< \dfrac{1}{2}\)
