由已知利用递推公式sn-sn-1=an,a1=s1可求通项,然后代入,利用错位相减即可求解数列的和
【解析】
∵Sn=(n2+n)﹒2n,
∴Sn-1=[(n-1)2+(n-1)]﹒2n-1,(n≥2)
两式相减可得,sn-sn-1=(n2+n)﹒2n-[(n-1)2+(n-1)]﹒2n-1,
=2n-1•(n2+3n)(n≥2)
n=1时,a1=s1=4适合上式
∴an=2n-1•(n2+3n)
∴=(n+3)•2n-1
∴sn=4•2+5•2+…+(n+3)•2n-1
2sn=4•2+5•21+…+(n+2)•2n-1+(n+3)•2n
两式相减可得,-sn=4+2+22+…+2n-1-(n+3)•2n
=4+-(n+3)•2n
=4+2n-2-(n+3)•2n
=2-(n+2)•2n
∴Sn=(n+2)•2n-2
故选C.