Question

[Nguyễn Phương Anh]

Tính:

a)  A = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) +...+ \(\dfrac{1}{998.999}\) + \(\dfrac{1}{999.1000}\)

b) B = \(\dfrac{1}{1.6}\) + \(\dfrac{1}{6.11}\) + \(\dfrac{1}{11.16}\) +...+ \(\dfrac{1}{495.500}\)

c) C = \(\dfrac{1}{1.2.3}\) + \(\dfrac{1}{2.3.4}\) + \(\dfrac{1}{3.4.5}\) +...+ \(\dfrac{1}{998.999.1000}\)

(Mong mn giúp ạ)

Answer 1

[Akai Haruma]

a.

$A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{1000-999}{999.1000}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}$

$=1-\frac{1}{1000}=\frac{999}{1000}$

Answer 2

[Akai Haruma]

b.

$5B=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+....+\frac{5}{495.500}$

$=\frac{6-1}{1.6}+\frac{11-6}{6.11}+\frac{16-11}{11.16}+....+\frac{500-495}{495.500}$

$=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{495}-\frac{1}{500}$

$=1-\frac{1}{500}=\frac{499}{500}$

$\Rightarrow B=\frac{499}{500}: 5= \frac{499}{2500}$