问题
填空题
设等差数列{an}的公差d是2，前n项的和为Sn，则
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答案
由公差d=2，得到an=a1+2（n-1）=2n+a1-2，Sn=na1+
×2=n2+n（a1-1）n(n-1) 2
则lim n→∞
=
-n2a 2n Sn lim n→∞
=3n2+4(a1-2)n+(a1-2)2 n2+n(a1-1) lim n→∞
=33+
+4(a1-2) n (a1-2)2 n2 1+ a1-1 n
故答案为3．