由a1=3,an+1=an+p·3n,得a2=3+3p,a3=a2+9p=3+12p. ∵a1,a2+6,a3成等差数列,∴a1+a3=2(a2+6),即3+3+12p=2(3+3p+6),得p=2. 依题意知,an+1=an+2×3n, 当n≥2时,a2-a1=2×31,a3-a2=2×32,…,an-an-1=2×3n-1. 等号两边分别相加得an-a1=2(31+32+…+3n-1)=2×=3n-3, ∴an-a1=3n-3,∴an=3n(n≥2). 又a1=3适合上式,故an=3n. (2)证明:∵an=3n,∴bn=. ∵bn+1-bn=-= (n∈N*). 若-2n2+2n+1<0,则n>, 即当n≥2时,有bn+1<bn. 又因为b1=,b2<.故bn≤ |