假设公差为d的等差数列｛an｝满足条件,则有
a1+2a2+3a3+……+nan
=a1+2(a1+d)+3(a1+2d)+……+n[a1+(n-1)d]
=a1(1+2+3+……+n)+d[2*1+3*2+4*3+……+n*(n-1)]
=a1*(n+1)n/2+d[2²-2+3²-3+4²-4+……+n²-n]
=a1*(n+1)n/2+d[(1²+2²+3²+4²+……+n²)-(1+2+3+4+……+n)]
=a1*(n+1)n/2+d[n(n+1)(2n+1)/6-(n+1)n/2]
=[n(n+1)/2]*[a1+d(2n-2)/3]
=n(n+1)(n+2)
a1+(2n-2)d/3=2(n+2)
(2d-6)n+3a1-2d-12=0
因为此式对任意自然数n均成立
所以2d-6=0
且3a1-2d-12=0
解得d=3,a1=6
所以an=a1+(n-1)d=6+3(n-1)=3n+3
即所求数列为an=3n+3