Question

已知数列an的前n项和为Sn,且Sn=n(n+1）（n属于N*)求数列an的通项公式；（2）若数列

已知数列an的前n项和为Sn,且Sn=n(n+1）（n属于N*)求数列an的通项公式；（2）若数列bn满足:an=b1/(3+1)+b2/(3^2+1）+b3/(3^3+1)求数列｛bn｝的通项公式

Answer 1

(1)
Sn=n(n+1)
n=1 , a1=2
an = Sn-S(n-1)
= 2n
(2)
an=b1/(3+1)+b2/(3^2+1）+...+bn/(3^n+1) (1)
a(n-1)=b1/(3+1)+b2/(3^2+1）+...+b(n-1)/(3^(n-1)+1) (2)
(1)-(2)
an - a(n-1) = bn/(3^n+1)
2n-2(n-1)= bn/(3^n+1)
bn = 2(3^n+1)