Question

Tính :

A 1/10×11 +1/11×12+1/12×13+...+1/99×100B 1/1×3+1/3×5+1/5×7+...+1/97×99

Answer 1

A=\(\frac{1}{10.11}\)+\(\frac{1}{11.12}\)+...+\(\frac{1}{99.100}\)

⇒A=\(\frac{1}{10}\)-\(\frac{1}{11}\)+\(\frac{1}{11}\)-\(\frac{1}{12}\)+...+\(\frac{1}{99}\)-\(\frac{1}{100}\)

⇒A=\(\frac{1}{10}\)-\(\frac{1}{100}\)

⇒A=\(\frac{9}{100}\)

Vậy A=\(\frac{9}{100}\)

B=\(\frac{1}{1.3}\)+\(\frac{1}{3.5}\)+...+\(\frac{1}{97.99}\)

=\(\frac{1}{2}\).\((1-\frac{1}{3})\)+\(\frac{1}{2}.(\frac{1}{3}-\frac{1}{5})\)+...+\(\frac{1}{2}.(\frac{1}{97}-\frac{1}{99})\)

=\(\frac{1}{2}.(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99})\)

=\(\frac{1}{2}.\frac{98}{99}\)

=\(\frac{49}{99}\)

Vậy B=\(\frac{49}{99}\)