Question

[Army of bts]

B1 : Tính giá trị biểu thức :

S = \(\frac{2}{1.3}\)-  \(\frac{4}{3.5}\)+  \(\frac{6}{5.7}\)-  \(\frac{8}{7.9}\)+  ...  -   \(\frac{96}{95.97}\)+  \(\frac{98}{97.99}\)

B2 : Chứng minh rằng A = \(\frac{1}{4}\)+  \(\frac{1}{16}\)+  \(\frac{1}{36}\)+  \(\frac{1}{64}\)+  \(\frac{1}{100}\)+  \(\frac{1}{144}\)+  \(\frac{1}{196}\)+  \(\frac{1}{256}\)+  \(\frac{1}{324}\)<  \(\frac{1}{2}\)

Answer 1

[Dũng Lê Trí]

B2 : \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{114}+\frac{1}{196}+\frac{1}{256}+\frac{1}{324}\)

\(=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{18^2}\)

\(\frac{1}{2^2}< \frac{1}{1\cdot2}\)

\(\frac{1}{4^2}< \frac{1}{2\cdot4}\)

\(\frac{1}{6^2}< \frac{1}{4\cdot6}\)

...

\(\frac{1}{18}< \frac{1}{16\cdot18}\)

\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{18^2}< \frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{16}-\frac{1}{18}\right)\)

\(\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{18^2}< \frac{1}{2}< \frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{18}\right)\)