Question

求解啊，数学计算题，要过程

3/（2+4+6+……+1004）+3/（2+4+6+……+1006）+3/(2+4+6+……+1008)+……3/(2+4+6+……+2006)=?
还有图中的问题

Answer 1

3/（2+4+6+……+1004）+3/（2+4+6+……+1006）+3/(2+4+6+……+1008)+……3/(2+4+6+……+2006)
=3/[(2+1004)*502/2]+3/[(2+1006)*503/2]+......+3/[(2+2006)*1003/2]
=3/[(1+502)*502]+3/[(1+503)*503]+......+3/[(1+1003)*1003]
=3/(503*502)+3/((504*503)+......+3/(1004*1003)
=3*[1/(503*502)+1/(504*503)+......+1/(1004*1003)]
=3*[(1/502-1/503)+(1/503-1/504)+......(1/1003-1/1004)]
=3*(1/502-1/503+1/503-1/504+......1/1002-1/1003+1/1003-1/1004)
=3*(1/502-1/1004)
=3*(1/1004)
=3/1004

追问

还有呢，你就回答了一点

Answer 2

3/（2+4+6+……+1004）+3/（2+4+6+……+1006）+3/(2+4+6+……+1008)+……3/(2+4+6+……+2006)=3/2*(2/502*503+2/503*504+2/504*505+...........2/1003*1004)
=3*(1/502-1/503+1/503-1/504+1/504-1/505+........+1/1003-1/1004)
=3*(2/1004-1/1004)
3/1004

追问

还有呢，你就回答了一点

Answer 3

等差数列 (首项+末项)*项数/2