Question

tính a) A= \(\frac{1}{10}\)+ \(\frac{1}{40}\)+ \(\frac{1}{88}\)+ \(\frac{1}{154}\)+ \(\frac{1}{238}\)+ \(\frac{1}{340}\)

b) so sánh 2004¹⁰+2004⁹ và 2005¹⁰

Answer 1

a, \(A=\frac{1}{10}+\frac{1}{40}+...+\frac{1}{340}\)

\(\Leftrightarrow A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{17.20}\)

\(\Leftrightarrow A=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+....+\frac{3}{17.20}\right)\)

\(\Leftrightarrow A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(\Leftrightarrow A=\frac{1}{6}-\frac{1}{60}=\frac{3}{20}\)

b,  \(2004^{10}+2004^9=2004^9\left(2014+1\right)=2014^9+2005\)

\(2015^{10}=2015^9.2015\)

-Vậy: \(2004^{10}+2004^9< 2005^{10}\)