Question

等差数列an的前n项和为sn,a1=1+√2 ,s3=9+3√2求数列{an}的通项an与前n项和Sn

Answer 1

S3=(a1+a3)*3/2=(a1+a1+2d)*3/2=2(a1+d)*3/2=2a2*3/2=3a2=9+3√2

所以a2=3+√2

d=a2-a1=2

所以an=a1+2(n-1)=√2-1+2n

Sn=(a1+an)*n/2=(2√2+2n)*n/2=n^2+2√2n