Question

A=1/2^2+1/4^2+1/6^2+...+1/14^2 hãy so sánh với  1/2

3 k

Answer 1

Ta co:\(A=\frac{1}{2.2}+\frac{1}{4.4}+\frac{1}{6.6}+...+\frac{1}{14.14}< \frac{2}{2.4}+\frac{2}{4.6}+\frac{1}{6.8}+...+\frac{2}{14.16}\left(1\right)\)

\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{14.16}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{14}-\frac{1}{16}\)

\(=\frac{1}{2}-\frac{1}{16}=\frac{7}{16}< \frac{8}{16}=\frac{1}{2}\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow A< \frac{1}{2}\)

V...

Answer 2

cho dong dau tien la dau =,chu ko phai dau < nhe

Answer 3

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