Question

3. Tính tổng

a) \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{9.10}\)

b) \(\dfrac{1}{100}\)-\(\dfrac{1}{100.99}\)-\(\dfrac{1}{99.98}\)-\(\dfrac{1}{98.97}\)-...-\(\dfrac{1}{3.2}\)-\(\dfrac{1}{2.1}\)

Giúp mình với

Answer 1

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{9.10}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{9}-\dfrac{1}{10}\)

\(A=1-\dfrac{1}{10}=\dfrac{9}{10}\)

\(B=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-.....-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(B=\dfrac{1}{100}-\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}+\dfrac{1}{97}-.....+\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{2}-1\)\(B=0-1=-1\)