解方程x²-﹙3k+2k﹚x+3k×2k=0
得x1=2k,x2=3k
∵a2k-1≤a2k
依题意：a(2k-1)=2k,a(2k)=3k
还问什么?数列﹛an﹜的前2n项和S2nk=1,a1=2,a2=3k=2,a3=4,a4=6k=3,a5=6,a6=9a(2k+1)-a(2k-1)=2(k+1)-2k=2a(2k+2)-a(2k)=3∴{a(2k-1)}为等差数列，｛a(2k)}为等差数列∴S2n=[a1+a3+a5+......+a(2n-1)]+[a2+a4+....+a2n] =[2+4+6+......+2n]+[3+6+9+.....+3n] =5(1+2+.......+n) =5/2*(n+1)n