Question

已知数列{an}满足2a(n+1)=an+a(n+2)(n∈N*),它的前n项和为sn,且a3=5,s6=36,

求数列an的通项公式。要过程谢了

Answer 1

由2a(n+1)=an+a(n+2)知数列为等差数列，由a3=a1+2d=5,s6=6a1+1/2*6*5d=36解得a1=1,d=2,所以通项为an=a1+(n-1)d=2n-1,前n项和为sn=na+1/2n(n-)d=n^2