Chứng minh:
a, A = 1 + \(\frac{1}{2^2}\)+ \(\frac{1}{3^2}\)+ \(\frac{1}{4^2}\)+ ... + \(\frac{1}{100^2}\)< 2
b, B = 1 + \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)+ .... + \(\frac{1}{63}\)< 6
c, C = \(\frac{1}{2}\). \(\frac{3}{4}\). \(\frac{5}{6}\). \(\frac{7}{8}\).... \(\frac{9999}{10000}\)< \(\frac{1}{100}\)
GIÚP MÌNH VỚI NHÉ, MÌNH ĐANG CẦN GẤP LẮM
\(a)\) Ta có :
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(............\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\)\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(\Rightarrow\)\(A< 1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\)\(A< 1+1-\frac{1}{100}\)
\(\Rightarrow\)\(A< 2-\frac{1}{100}< 2\)
\(\Rightarrow\)\(A< 2\) ( đpcm )
Vậy \(A< 2\)
Chúc bạn học tốt ~
