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 với \(\frac{1}{4}\).

Answer 1

Ta có:

  B=\(\frac{4^2-2^2}{2^2\times4^2}+\frac{6^2-4^2}{4^2\times6^2}+...+\frac{98^2-96^2}{96^2\times98^2}+\frac{100^2-98^2}{98^2\times100^2}\)

   =\(\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)

  = \(\frac{1}{4}-\frac{1}{100^2}< \frac{1}{4}\) 

Answer 2

Ai làm nhanh và đúng nhất thì mình k cho nhé <3

Answer 3

Sao tánh ra dc thành 1/2^2-1/4^2 vậy

Answer 4

Lê Anh Quân sai r

\(2^2.4^2=4.6\)và \(6-4\ne12\)

Answer 5

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

  =\(\frac{12}{2^2.4^2}+\frac{20}{4^2.6^2}+...+\frac{388}{96^2.98^2}+\frac{396}{98^2.100^2}\)

  =\(\frac{4^2-2^2}{2^2.4^2}+\frac{6^2-4^2}{4^2.6^2}+...+\frac{98^2-96^2}{96^2.98^2}+\frac{100^2-98^2}{98^2.100^2}\)

  =\(\frac{1}{2^2.4^2}+\frac{1}{4^2.6^2}+...+\frac{1}{96^2.98^2}+\frac{1}{98^2.100^2}\)

  =\(\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)

  =\(\frac{1}{2^2}-\frac{1}{100^2}\)

  =\(\frac{1}{4}-\frac{1}{100^2}\)

Vì B=\(\frac{1}{4}-\frac{1}{100^2}\)mà \(\frac{1}{4}\)=\(\frac{1}{4}-\frac{0}{100^2}\)

Nên B<\(\frac{1}{4}\)