A=2/3+2/3^2+...+2/3^10
=>3A=2+2/3+...+2/3^9
=>2A=2-2/3^10
=>A=1-1/3^10<1
A=2/3+2/3^2+...+2/3^10
=>3A=2+2/3+...+2/3^9
=>2A=2-2/3^10
=>A=1-1/3^10<1
Cho A = \(\dfrac{1}{2}\) + \(\dfrac{1}{2^{2}}\)+ \(\dfrac{1}{2^{3}}\)+ \(\dfrac{1}{2^{4}}\) + ...+ \(\dfrac{1}{2^{10}}\)
Chứng tỏ rằng A + \(\dfrac{1}{2^{10}}\)= 1
Chứng tỏ rằng: A= 2+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}^{ }\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{100^2}\)<3
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:\(\dfrac{1}{2^2}\dfrac{1}{3^2}\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{3}{4}\)
cho em lời giải chi tiết với ạ
Chứng tỏ rằng: \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)=2
\(\text{Bài 4. Chứng tỏ rằng:}\)
\(a\)) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}< 1\)
\(b\)) \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}>1\)
\(c\)) \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)
\(d\)) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}< 1\)
Chứng tỏ rằng : \(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{60^2}< \dfrac{4}{9}\)
chứng tỏ rằng:\(\dfrac{1}{2^3}+\dfrac{1}{4^2}+...+\dfrac{1}{60^2}< \dfrac{1}{9}\)
chứng tỏ rằng: \(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{60^2}< \dfrac{1}{9}\)