Question

Cho B = \(\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+..........+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)

Hãy so sánh B và \(\frac{1}{4}\)

Answer 1

Ta có :

\(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)

\(=\frac{12}{4.16}+\frac{20}{16.36}+...+\frac{388}{9216.9604}+\frac{396}{9604.10000}\)

\(=\frac{1}{4}-\frac{1}{16}+\frac{1}{16}-\frac{1}{36}+...+\frac{1}{9604}-\frac{1}{10000}\)

\(=\frac{1}{4}-\frac{1}{10000}< \frac{1}{4}\)

\(\Leftrightarrow B< \frac{1}{4}\)

Answer 2

B=\(\frac{12}{4.16}\)+\(\frac{20}{16.36}\)+...+\(\frac{396}{9604.10000}\)

Ta có:\(\frac{12}{4.16}\)=\(\frac{1}{4}\)-\(\frac{1}{16}\)

         \(\frac{20}{16.36}\)=\(\frac{1}{16}\)-\(\frac{1}{36}\)

            ...

Khi đó:B=\(\frac{1}{4}\)-\(\frac{1}{16}\)+\(\frac{1}{16}\)-\(\frac{1}{36}\)+...+\(\frac{1}{9604}\)-\(\frac{1}{10000}\)=\(\frac{1}{4}\)-\(\frac{1}{10000}\)<\(\frac{1}{4}\)

Vậy: B<\(\frac{1}{4}\)