Question

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

Ai nhanh mik tick cho

Answer 1

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

\(B=\frac{4^2-2^2}{\left(2.4\right)^2}+\frac{6^2-4^2}{\left(4.6\right)^2}+..........+\frac{98^2-96^2}{\left(96.98\right)^2}+\frac{100^2-98^2}{\left(98.100\right)^2}\)

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

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

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

\(B=\frac{2500}{10000}-\frac{1}{10000}\)

\(B=\frac{2499}{10000}\)

Vậy B = \(\frac{2499}{10000}\)