(1)bn＝2n-1 (n∈N*)Sn＝(n+1)bn＝2n^2+n-1①故S(n-1)＝2(n-1)^2+(n-1)-1＝2n^2-3n②(n≥2,n∈N*)①-②得an＝4n-1(n≥2,n∈N*)当n＝1,S1＝a1＝2(1)^2+1-1＝2而a1＝1×4-1＝3≠2故an＝2,n＝1an=4n-1,n≥2,n∈N*(2)当n...能讲解一下裂项相消法么？如本题的1/[(4n-1)(4n+3)]=1/4*[1/(4n-1)-1/(4n+3)]，Tn=c1+[c2+……+cn] c2=1/4*(1/7-1/11)c3=1/4*(1/11-1/15)c2+c3=1/4*（1/7-1/11+1/11-1/15）中间两项就可以消去这个就叫裂项相消法