Question

a1=3,an+an-1=2+ n(3n+1)/an-an-1 (其中,an-1 表示an的前一项）,求通向

Answer 1

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