Question

已知数列{an}的前n项和为Sn，求an

Answer 1

3的等比数列
（2）
an=(2/3)[∑n^2+3∑n+∑2]
=(2/6
∴an=(2/3)(n+2)(n+1)
∴bn=an/3)(n+1)
∴bn-b(n-1)=2/冲扮a(n-2)=(n+1)/a1=4/2)n(n+1)+2n]
=(2n/，o(∩_∩)o.;a1=(n+2)/念判空(n-1)
.;n
a(n-1)/.(1)∵a1=4;3*2)=(n+2)(n+1)/.;n*(1/3)(n+2)(n+1)
=(2/.
a2/3)[(1/2
∴an/(n+1)=(2/9)(n^2+6n+11);3)(n^2+3n+2)
sn=(2/3)(n+2)
∴b(n-1)=(2/3
∴{bn}是公比为2/6)n(n+1)(2n+1)+(3/，希望对你仔瞎有帮助,(n+1)an+1=(n+3)an
∴an/a(n-1)=(n+2)/