Ta có:

1+\(\dfrac{1}{1+2}\)=  1+\(\dfrac{1}{\dfrac{2\left(2+1\right)}{2}}\)= 1+ \(\dfrac{2}{2.3}\)

1+\(\dfrac{1}{1+2+3}\)= 1+ \(\dfrac{1}{\dfrac{3\left(3+1\right)}{2}}\)=  1+ \(\dfrac{2}{3.4}\)

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1+ \(\dfrac{1}{1+2+3+...+997}\)= 1+ \(\dfrac{1}{\dfrac{997\left(997+1\right)}{2}}\)=  1+\(\dfrac{2}{997.998}\)

=>  A = 1+\(\dfrac{2}{2.3}\)+1+\(\dfrac{2}{3.4}\)+......+1+\(\dfrac{2}{997.998}\)

          =   (1+1+....+1) +   2(\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+.....+\(\dfrac{1}{997.998}\))

          =   1.996 +  2( \(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+....+\(\dfrac{1}{997}\)-\(\dfrac{1}{998}\))

          =   996 +   2(\(\dfrac{1}{2}\)-\(\dfrac{1}{998}\))

Chỗ này bạn tự tính nốt ra nha!!( Chú ý: Dấu chấm chính là dấu nhân đấy không phải dấu phẩy đâu)