Ta có: \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{14.17}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{14}-\frac{1}{17}\)
\(=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)\(< \frac{17}{34}=\frac{1}{2}\)
\(\Rightarrow\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{14.17}< \frac{1}{2}\)
Vậy:..........................................(đpcm)
1. tính
a) A=\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
b) B=\(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)
c) C=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
d) D=\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.98}\)
a) \(c=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
b) \(d=\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+...+\frac{6}{87.90}\)
c) \(e=\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+...+\frac{3}{197.200}\)
Mik đang cần gấp
\(\frac{1}{5.8}+\frac{1}{8.11}+......+\frac{1}{302.305}\)
Tính nhanh:
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.11}+\frac{3}{11.14}+\frac{3}{14.17}\)
chứng tỏ rằng:
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{99^2}+\frac{1}{100^2}< \frac{3}{4}\)
giúp mình nhé mình đang cần gấp!
Cho S = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2012}}+\frac{1}{2^{2013}}\) Chứng tỏ S < 1
ChoA=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}....+\frac{1}{49}-\frac{1}{50}\)
B=\(\frac{1}{25}+\frac{1}{26}+\frac{1}{27}.....+\frac{1}{50}\)
Chứng tỏ rằng A=B
Bài 1: Chứng minh rằng: \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
Bài 2: Cho \(n\in N;n>1\). Chứng minh rằng: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{\left(n-1\right)^2}+\frac{1}{n^2}\notin N\)
Chứng Minh Rằng
a) \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b) \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+.....+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
