THAM KHẢO NHA 

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\(A=\dfrac{1}{4}+\dfrac{1}{16}+...+\dfrac{1}{576}\)

=1/2^2(1+1/2^2+...+1/12^2)

1/2^2+1/3^2+...+1/12^2<1-1/2+1/2-1/3+...+1/11-1/12=11/12

=>1+1/2^2+1/3^2+...+1/12^2<1+11/12=23/12

=>A<23/12*1/4=23/48<23/46=1/2

A = \(\dfrac{1}{4}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{36}\) +...+ \(\dfrac{1}{196}\) 

A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\)+...+ \(\dfrac{1}{14^2}\)

A = \(\dfrac{1}{\left(1.2\right)^2}\) + \(\dfrac{1}{\left(2.2\right)^2}\) + \(\dfrac{1}{\left(2.3\right)^2}\)+...+ \(\dfrac{1}{\left(2.7\right)^2}\)

A = \(\dfrac{1}{1^2.2^2}\) + \(\dfrac{1}{2^2.2^2}\)+ \(\dfrac{1}{2^2.3^2}\)+...+ \(\dfrac{1}{2^2.7^2}\)

A = \(\dfrac{1}{2^2}\) \(\times\)( \(\dfrac{1}{1}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\)+...+ \(\dfrac{1}{7^2}\))

Vì \(\dfrac{1}{2}>\dfrac{1}{3}>\dfrac{1}{4}>\dfrac{1}{5}\) \(>\)\(\dfrac{1}{6}>\dfrac{1}{7}\) 

⇒ \(\dfrac{1}{2.2}\)+\(\dfrac{1}{3.3}\)+\(\dfrac{1}{4.4}\)+\(\dfrac{1}{5.5}\)+\(\dfrac{1}{6.6}\)+\(\dfrac{1}{7.7}\) < \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{4.5}\)+\(\dfrac{1}{5.6}\)+\(\dfrac{1}{6.7}\)

⇒ A < \(\dfrac{1}{2^2}\) \(\times\) ( 1 + \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\))

⇒ A < \(\dfrac{1}{4}\) \(\times\) ( 2 - \(\dfrac{1}{7}\))

⇒ A < \(\dfrac{1}{2}\) - \(\dfrac{1}{28}\) < \(\dfrac{1}{2}\)

⇒ A < \(\dfrac{1}{2}\) ( đpcm)

Giúp đi

dpcm là gì vậy các bồ

dpcm là điều phải chứng minh nha

Ta có :     \(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)

                 \(\frac{1}{16}< \frac{1}{4}-\frac{4}{8}\)

                \(\frac{1}{36}< \frac{1}{8}-\frac{1}{12}\)

                \(\frac{1}{64}< \frac{1}{12}-\frac{1}{16}\)

                 \(\frac{1}{100}< \frac{1}{16}-\frac{1}{20}\)

                   \(\frac{1}{144}< \frac{1}{20}-\frac{1}{24}\)

                 \(\frac{1}{196}< \frac{1}{24}-\frac{1}{28}\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{28}\)

             \(=\frac{1}{2}-\frac{1}{28}< \frac{1}{2}\)

Vậy A<1/12