等差數(shù)列列｛an｝的前n項(xiàng)和為Sn,公差為2,則lim(n→∞)Sn/an^2=?
等差數(shù)列列｛an｝的前n項(xiàng)和為Sn,公差為2,則lim(n→∞)Sn/an^2=?
數(shù)學(xué)人氣：297 ℃時(shí)間：2019-10-23 04:21:19
優(yōu)質(zhì)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->無(wú)窮大){s(n)/[a(n)]^2}=lim(n->無(wú)窮...
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