等差数列{an}的公差为d(d≠0),它的前n项的和为Sn;等比数列{bn}的公比为q(q的绝对值>1),它的前n项和为Bn求lim(Sn、n*an-Bn/bn)lim(Sn/(n*an)-Bn/bn)
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![等差数列{an}的公差为d(d≠0),它的前n项的和为Sn;等比数列{bn}的公比为q(q的绝对值>1),它的前n项和为Bn求lim(Sn、n*an-Bn/bn)lim(Sn/(n*an)-Bn/bn)](/uploads/image/z/13919973-69-3.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%85%AC%E5%B7%AE%E4%B8%BAd%28d%E2%89%A00%29%2C%E5%AE%83%E7%9A%84%E5%89%8Dn%E9%A1%B9%E7%9A%84%E5%92%8C%E4%B8%BASn%3B%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%7Bbn%7D%E7%9A%84%E5%85%AC%E6%AF%94%E4%B8%BAq%28q%E7%9A%84%E7%BB%9D%E5%AF%B9%E5%80%BC%3E1%EF%BC%89%2C%E5%AE%83%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BABn%E6%B1%82lim%EF%BC%88Sn%E3%80%81n%2Aan-Bn%2Fbn%29lim%28Sn%2F%28n%2Aan%29-Bn%2Fbn%29)
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等差数列{an}的公差为d(d≠0),它的前n项的和为Sn;等比数列{bn}的公比为q(q的绝对值>1),它的前n项和为Bn求lim(Sn、n*an-Bn/bn)lim(Sn/(n*an)-Bn/bn)
等差数列{an}的公差为d(d≠0),它的前n项的和为Sn;等比数列{bn}的公比为q(q的绝对值>1),它的前n项和为Bn
求lim(Sn、n*an-Bn/bn)
lim(Sn/(n*an)-Bn/bn)
等差数列{an}的公差为d(d≠0),它的前n项的和为Sn;等比数列{bn}的公比为q(q的绝对值>1),它的前n项和为Bn求lim(Sn、n*an-Bn/bn)lim(Sn/(n*an)-Bn/bn)
用等差和等比前n项和公式把Sn和Bn代换后化简后:
=lim(2a1+(n-1)d)/(2a1+2(n-1)d)—lim(1-q^n)/(q^n-1(1-q))
=1/2-1
=-1/2
lim()里面什么,看不懂
根据题意得Sn=na1+n(n-1)d/2,Bn=b1[1-q^(n-1)]/(1-q),
an=a1+(n-1)d,bn=b1q^(n-1)
所以lim(Sn/(n*an)-Bn/bn)=lim{[na1+n(n-1)d/2]/n[a1+(n-1)d]-[b1(1-q^(n-1))/(1-q)]/b1q^(n-1)}
=1/2-(-1)=3/2
你写的啥么,根本看不懂
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