b: \(B=\frac15+\frac{2}{5^2}+\frac{3}{5^3}+\cdots+\frac{50}{5^{50}}\)
=>\(5B=1+\frac25+\frac{3}{5^2}+\cdots+\frac{50}{5^{49}}\)
=>\(5B-B=1+\frac25+\frac{3}{5^2}+\cdots+\frac{50}{5^{49}}-\frac15-\frac{2}{5^2}-\frac{3}{5^3}-\cdots-\frac{49}{5^{49}}-\frac{50}{5^{50}}\)
=>\(4B=1+\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{49}}-\frac{50}{5^{50}}\)
Đặt \(E=1+\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{49}}\)
=>\(5E=5+1+\frac15+\cdots+\frac{1}{5^{48}}\)
=>\(5E-E=5+1+\frac15+\cdots+\frac{1}{5^{48}}-1-\frac15-\frac{1}{5^2}-\cdots-\frac{1}{5^{49}}\)
=>\(4E=5-\frac{1}{5^{49}}=\frac{5^{50}-1}{5^{49}}\)
=>\(E=\frac{5^{50}-1}{4\cdot5^{49}}\)
\(4B=E-\frac{50}{5^{50}}=\frac{5^{50}-1}{4\cdot5^{49}}-\frac{50}{5^{50}}=\frac{5^{51}-5}{4\cdot5^{50}}-\frac{200}{4\cdot5^{50}}=\frac{5^{51}-205}{5^{50}\cdot4}\)
=>\(4B<\frac{5^{51}}{5^{50}\cdot4}=\frac54\)
=>\(B<\frac{5}{16}\)
c: \(C=\frac{1}{5^2}+\frac{1}{10^2}+\frac{1}{15^2}+...+\frac{1}{100^2}\)
\(=\frac{1}{5^2}\left(1+\frac{1}{2^2}+\cdots+\frac{1}{20^2}\right)\)
Đặt \(F=1+\frac{1}{2^2}+\cdots+\frac{1}{20^2}\)
\(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)
\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)
...
\(\frac{1}{20^2}<\frac{1}{19\cdot20}=\frac{1}{19}-\frac{1}{20}\)
Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{20^2}<1-\frac12+\frac12-\frac13+\cdots+\frac{1}{19}-\frac{1}{20}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{20^2}<1\)
=>\(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{20^2}<1+1=2\)
=>\(C=\frac{1}{5^2}\cdot F=\frac{1}{25}\cdot F<\frac{1}{25}\cdot2=\frac{2}{25}\)
=>ĐPCM
d: \(D=7+\frac{343}{14^2}+\frac{343}{21^2}+\cdots+\frac{343}{210^2}\)
\(=\frac{343}{7^2}+\frac{343}{14^2}+\cdots+\frac{343}{210^2}\)
\(=\frac{343}{7^2}\left(1+\frac{1}{2^2}+\cdots+\frac{1}{30^2}\right)\)
\(=7\left(1+\frac{1}{2^2}+\cdots+\frac{1}{30^2}\right)\)
Đặt \(G=1+\frac{1}{2^2}+\cdots+\frac{1}{30^2}\)
\(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)
\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)
....
\(\frac{1}{30^2}<\frac{1}{29\cdot30}=\frac{1}{29}-\frac{1}{30}\)
Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{30^2}<1-\frac12+\frac12-\frac13+\cdots+\frac{1}{29}-\frac{1}{30}=1-\frac{1}{30}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{30^2}<1\)
=>\(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{30^2}<1+1=2\)
=>G<2
\(D=7\cdot G<7\cdot2=14\)
=>ĐPCM
