高中数列问题求助——————

高中数列问题求助——————

an -a(n-1) =n(n+1)^2 -1
an -a(n-1) =n^3+2n^2+n-1

an -a(n-1) =n^3+2n^2+n-1
.................
a3-a2=3^3+2*3^2+3-1
a2-a1=2^3+2*2^2+2-1
以上等式相加得
an-a1=2^3+2*2^2+2-1+3^3+2*3^2+3-1+...........+n^3+2n^2+n-1
an-a1=2^3+3^3+......+n^3+2*2^2+2*3^2+........+2n^2+2+3+..........+n-1-1-..........-1
an-a1=1^3+2^3+3^3+......+n^3+2*1^2+2*2^2+2*3^2+........+2n^2+1+2+3+..........+n-(n-1)-4
an-3=[n(n+1)/2]^2+2*n(n+1)(2n+1)/6+n(n+1)/2-n-3
an=n^2(n+1)^2/4+n(n+1)(2n+1)/3+n(n+1)/2-n
an=n(n+1)/12[3n(n+1)+4(2n+1)+6]-n
an=n(n+1)/12[3n^2+3n+8n+4+6]-n
an=n(n+1)(3n^2+11n+10)/12-n

∵an+a(n-1)=2+ n(3n+1)/[an-a(n-1)]
∴a²n-a²(n-1)=2[an-a(n-1)]+ n(3n+1)
即(an-1)²-[a(n-1)-1]²=n(3n+1)=3n²+n,n≥2
令Sn为数列{(an-1)²-[a(n-1)-1]²}的前n项和，
则Sn=(an-1)²-(a1-1)²
=3(2²+3²+..+n²)+(2+3+...+n)=3n(n+1)(2n+1)/6+n(n+1)/2-3-1
=n(n+1)(2n+1)/2+n(n+1)/2-4=n(n+1)²-4
∵a1=3∴(an-1)²-2²=n(n+1)²-4
an=(√n)(n+1)+1,n≥2......这里需要证明一下an-1≥0
∵a1=3满足上式
∴an=(√n)(n+1)+1,n≥1