\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+....+\frac{19}{9^2.10^2}\)
\(A=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+....+\frac{19}{81.100}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+....+\frac{1}{81}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}< 1\)
\(\Rightarrow A< 1\text{(đpcm) }\)
Chướng minh rằng:
a, \(\frac{1}{1^2.2^2}\)+$\frac{5}{2^2.3^2}$+$\frac{5}{3^2.4^2}$+...+$\frac{5}{9^2.10^2}$ <1
b, \(\frac{1}{3}\)+\(\frac{2}{3^2}\)+$\frac{3}{3^3}$+$\frac{4}{3^4}$+...+$\frac{100}{3^100}$ <\(\frac{3}{4}\)
A=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)\(< \frac{3}{4}\)
B=\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\)
C=\(1+3+3^2+3^3+...+3^{100}\)
D=\(2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
Chứng tỏ rằng:
a) \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{17}{8^2.9^2}+\frac{19}{9^2.10^2}< 1\) 1
b) \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{99}{3^{99}}+\frac{100}{3^{100}}< \frac{3}{4}\)
Giúp mik với, sáng mai 8h00 mik cần gấp. Bạn nào nhanh với trình bày đầy đủ mik tick cho ~_~
Chướng minh rằng:
a, \(\frac{1}{1^2.2^2}\)+\(\frac{5}{2^2.3^2}\) +\(\frac{5}{3^2.4^2}+...+\frac{5}{9^2.10^2}\)<1
b, \(\frac{1}{3}\)+\(\frac{2}{3^2}\) +\(\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)<\(\frac{3}{4}\)
Bài 1: So sánh:
a) 5^255 và 2^512 ( 2 cách)
b) 8^12 và 12^8
Bài 2: Chứng minh rằng:
a) A = \(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{99}<1\)
b) B = \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}<1\)
c) C = \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}<\frac{3}{4}\)
Chứng minh rằng: A=\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}<1\)
Chứng minh: \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}
Chứng minh rằng: \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}
