摘要:二项式定理:(1)掌握二项展开式的通项: (2)注意第r+1项二项式系数与第r+1系数的区别,
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已知{an}为等比数列,a1=1,a4=27.Sn为等差数列{bn}的前n项和,b1=3,S5=35.
(1)求{an}和{bn}的通项公式;
(2)设Tn=a1b1+a2b2+…+anbn,求Tn.
已知数列{an}的首项a1=5,前n项和为Sn,且Sn+1=2Sn+n+5(n∈N*).
(Ⅰ)设bn=an+1,求数列{bn}的通项公式;
(Ⅱ)求数列{an}的前n项和Sn.
已知数列{an}、{bn}满足a1=2,an-1=an(an+1-1),bn=an-1.
(Ⅰ)求数列{bn}的通项公式;
(Ⅱ)若数列{bn}的前n项和为Sn,设Tn=S2n-Sn,求证:Tn+1>Tn.