\(B=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....++\frac{1}{9}-\frac{1}{10}\)
\(B=1-\frac{1}{10}=\frac{9}{10}\)
\(C=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(C=1-\frac{1}{100}\)
\(C=\frac{99}{100}\)
\(D=\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{496.501}\)
\(D=\frac{1}{5}\cdot\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+.....+\frac{1}{496}-\frac{1}{501}\right)\)
\(D=\frac{1}{5}\cdot\left(1-\frac{1}{501}\right)=\frac{1}{5}\cdot\frac{500}{501}=\frac{100}{501}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^n}\)
= \(\frac{2-1}{2}+\frac{2-1}{2^2}+\frac{2-1}{2^3}+...+\frac{2-1}{2^n}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{2^3}+...+\frac{1}{2^{n-1}}-\frac{1}{2^n}\)
\(=1-\frac{1}{2^n}\)
=\(\frac{2^n-1}{2^n}\)
