Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ta có: B=1/4^2 +1/6^2+......+1/200^2
B= 1/2^2 x(1/2^2+1/3^2 +1/4^2+......+1/100^2)
B=1/4 x(A)
suy ra: A/B= A / 1/4 x A
A/B= 4
Ta có:
\(2^2<4^2\Rightarrow\frac{1}{2^2}>\frac{1}{4^2}\)
\(3^2<6^2\Rightarrow\frac{1}{3^2}>\frac{1}{6^2}\)
\(4^2<8^2\Rightarrow\frac{1}{4^2}<\frac{1}{8^2}\)
\(...\)
\(100^2<200^2\Rightarrow\frac{1}{100^2}>\frac{1}{200^2}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}>\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{200^2}\)
\(\Rightarrow A>B\)
\(B=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+....+\frac{1}{200^2}=\frac{1}{\left(2.2\right)^2}+\frac{1}{\left(2.3\right)^2}+\frac{1}{\left(2.4\right)^2}+...+\frac{1}{\left(2.100\right)^2}\)
\(B=\frac{1}{2^2.2^2}+\frac{1}{2^2.3^2}+\frac{1}{2^2.4^2}+...+\frac{1}{2^2.100^2}=\frac{1}{2^2}.\frac{1}{2^2}+\frac{1}{2^2}.\frac{1}{3^2}+\frac{1}{2^2}.\frac{1}{4^2}+...+\frac{1}{2^2}.\frac{1}{100^2}\)
\(B=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}\right)=\frac{1}{4}.A\)
\(\Rightarrow\frac{A}{B}=\frac{A}{\frac{1}{4}A}=\frac{A}{\frac{A}{4}}=A.\frac{4}{A}=4\)
