解:(I)由 an=2an﹣1+2n﹣1(n∈N+,且n≥2)得 a4=2a3+24﹣1=81,得a3=33, 同理,可得 a2=13,a1=5. (II)∵an=2an﹣1+2n﹣1, ∴ ﹣ = ﹣ =1, 故数列 是以2为首项,以1为公差的等差数列. (III)由(II)可得 =2+(n﹣1)×1, ∴an=(n+1)2n+1. ∴Sn=a1+a2+…+an=2×2+3×22+4×23+…+(n+1)×2n+n, 记Tn=2×2+3×22+4×23+…+(n+1)×2n, 则有2Tn=2×22+3×23+…+n×2n +(n+1)2n+1. 两式相减, 可得﹣Tn=2×2+22+23+…+2n﹣(n+1)2n+1=4+ ﹣(n+1)2n+1=﹣n·2n+1, 解得 Tn=n×2n+1,故 Sn=Tn+n=n×2n+1+n=n?(2n+1+1 ). |