问题
填空题
若数列{an} 满足:an=2n+1,则其前n 项和Sn=______.
答案
∵an=2n+1,
∴Sn=21+1+22+1+…+2n+1
=
+n=2(2n-1)+n2(1-2n) 1-2
故答案为:2(2n-1)+n
若数列{an} 满足:an=2n+1,则其前n 项和Sn=______.
∵an=2n+1,
∴Sn=21+1+22+1+…+2n+1
=
+n=2(2n-1)+n2(1-2n) 1-2
故答案为:2(2n-1)+n