Question

[Vũ Thị Ngọc Doan]

1/1+2 + 1/1+2+3 + 1/1+2+3+4+.....+1/1+2+3+.....+1009

Answer 1

[Nguyễn Thị Thương Hoài]

A = \(\dfrac{1}{1+2}\) + \(\dfrac{1}{1+2+3}\)+ \(\dfrac{1}{1+2+3+4}\)+...+\(\dfrac{1}{1+2+3+...+1009}\)

Ta có công thức: 

S = 1 + 2 + ...+ n = (n+1)\(\times\)n:2

Áp dụng công thức trên vào A ta có

A = \(\dfrac{1}{\left(2+1\right)\times2:2}\)+\(\dfrac{1}{\left(1+3\right)\times3:2}\)+...+\(\dfrac{1}{\left(1009+1\right)\times1019:2}\)

A = \(\dfrac{1}{2\times3:2}\)+\(\dfrac{1}{3\times4:2}\)+...+\(\dfrac{1}{1009\times1010:2}\)

A = \(\dfrac{2}{2\times3}\)+ \(\dfrac{2}{3\times4}\)+...+\(\dfrac{2}{1009\times1010}\)

A = 2 \(\times\)( \(\dfrac{1}{2\times3}\)+ \(\dfrac{1}{3\times4}\)+...+ \(\dfrac{1}{1009\times1010}\))

A = 2 \(\times\) ( \(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+...+ \(\dfrac{1}{1009}-\dfrac{1}{1010}\))

A = 2 \(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{1010}\))

A = 1 - \(\dfrac{2}{1010}\)

A = \(\dfrac{1008}{1010}\)

A = \(\dfrac{504}{505}\)