> 数学 >
等差数列列{an}的前n项和为Sn,公差为2,则lim(n→∞)Sn/an^2=?
人气:312 ℃ 时间:2019-10-23 04:21:19
解答
a(n)=a+(n-1)d,[a(n)]^2=[a+(n-1)d]^2=[nd+a-d]^2,s(n)=na+n(n-1)d/2=(d/2)n^2+n(a-d/2),s(n)/[a(n)]^2=[(d/2)n^2+n(a-d/2)]/[nd+a-d]^2=[d/2+(a-d/2)/n]/[d+(a-d)/n]^2,lim(n->无穷大){s(n)/[a(n)]^2}=lim(n->无穷...
推荐
猜你喜欢
© 2025 79432.Com All Rights Reserved.
电脑版|手机版