Question

Answer 1

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

\(B=\dfrac{12}{2^2.4^2}+\dfrac{20}{4^2.6^2}+...+\dfrac{396}{98^2.100^2}\)

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

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

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

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