Question

Giúp mk với nhé

Answer 1

\(\frac{1}{12}\)+\(\frac{1}{20}\)+\(\frac{1}{30}\)+...+\(\frac{1}{9702}\)=\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+\(\frac{1}{5.6}\)+...+\(\frac{1}{98.99}\)=\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{6}\)+...+\(\frac{1}{98}\)-\(\frac{1}{99}\)=\(\frac{1}{3}\)+(-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{6}\)+...+\(\frac{1}{98}\))-\(\frac{1}{99}\)=\(\frac{1}{3}\)+(-\(\frac{1}{99}\))=\(\frac{32}{99}\)

Answer 2

Answer 3

\(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9702}=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{98.99}\)

=\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)