a) Chứng minh rằng \(\frac{1}{3^2}\) + \(\frac{1}{4^2}\) + \(\frac{1}{5^2}\) + \(\frac{1}{6^2}\) + ... + \(\frac{1}{100^2}\) < \(\frac{1}{2}\)
b) Cho phân số \(\frac{a}{b}\) và \(\frac{a}{c}\) có b = a - c ( a,b \(\in\) Z , b \(\ne\) 0 , c \(\ne\) 0 ). Chứng tỏ rằng: \(\frac{a}{b}\) . \(\frac{a}{c}\) = \(\frac{a}{b}\) + \(\frac{a}{c}\)và cho VD minh họa
a) \(A=\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(\Rightarrow A< \frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
b) b = a - c => b + c = a
\(\left\{{}\begin{matrix}\frac{a}{b}\cdot\frac{a}{c}=\frac{a^2}{bc}\\\frac{a}{b}+\frac{a}{c}=\frac{ac+ab}{bc}=\frac{a\left(b+c\right)}{bc}=\frac{a^2}{bc}\end{matrix}\right.\)
\(\Rightarrow\frac{a}{b}\cdot\frac{a}{c}=\frac{a}{b}+\frac{a}{c}\)