已知等差数列an=2n+1,求和s1+s2+s3.sn
2个回答

等差数列an=2n+1

则a1=3,d=a2-a1=2

所以Sn=3n+n(n-1)*2/2=n²+2n

所以S1+S2+S3+……+Sn

=(1²+2*1)+(2²+2*2)+(3²+2*3)+(4²+2*4)+……+(n²-2n)

=(1²+2²+3²+4²+……+n²)+2(1+2+3+4+……+n)

=n(n+1)(2n+1)/6+n(n+1)

=n(n+1)(2n+7)/6