(1)分a=1,a≠1两种情况求解,当a=1时为等差数列易求;当a≠1时利用错位相减法即可求得;
(2)其通项为an=n(n+2)=n2+2n,根据通项对数列各项进行分组求和,再运用公式即可求得;
【解析】
(1)当a=1时,Sn=1+2+3+…+n=;
当a≠1时,Sn=a+2a2+3a3+…+nan,①
aSn=a2+2a3+3a4+…+nan+1,②
①-②得,(1-a)Sn=a+a2+a3+a4+…+an-nan+1=-nan+1,
所以Sn=.
所以当a=1时,Sn=;当a≠1时,Sn=.
(2)数列通项an=n(n+2)=n2+2n,
则Sn=1×3+2×4+3×5+4×6+…+n(n+2)
=(12+2×1)+(22+2×2)+(32+2×3)+(42+2×4)+…+(n2+2n)
=(12+22+32+…+n2)+2(1+2+3+4+…+n)
=+2×=n2+n.