问题
填空题
等差数列{an}满足
|
答案
设等差数列的公差为d，首项为a1
∴Sn=na1+
n(n-1)d1 2
∴lim n→∞
=Sn 2n2 lim n→∞
=na1+
n(n-1)d1 2 2n2 lim n→∞
+a1 n
d(1-1 2
) 1 n 2
=
=1d 4
∴d=4
故答案为：an=4n
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等差数列{an}满足
|
设等差数列的公差为d，首项为a1
∴Sn=na1+
n(n-1)d1 2
∴lim n→∞
=Sn 2n2 lim n→∞
=na1+
n(n-1)d1 2 2n2 lim n→∞
+a1 n
d(1-1 2
) 1 n 2
=
=1d 4
∴d=4
故答案为：an=4n