倒序相加法推导前n项和公式:
Sn=a1+a2+a3 +·····+an
=a1+(a1+d)+(a1+2d)+······+[a1+(n-1)d] ①
Sn=an+an-1+an-2+······+a1
=an+(an-d)+(an-2d)+······+[an-(n-1)d] ②
由①+②得2Sn=(a1+an)+(a1+an)+······+(a1+an)(n个)=n(a1+an)
∴Sn=n(a1+an)÷2
等差数列的前n项和等于首末两项的和与项数乘积的一半:
Sn=n(a1+an)÷2=na1+n(n-1)d÷2
Sn=dn2÷2+n(a1-d÷2)
亦可得a1=2sn÷n-an=[sn-n(n-1)d÷2]÷n
an=2sn÷n-a1
有趣的是S2n-1=(2n-1)an,S2n+1=(2n+1)an+1
设数列{an}的首项a1=1,其前n项和Sn满足: 3tSn-(2t+3)Sn-1=3t(t>0,n=2,3,…).
(Ⅰ)求证:数列{an}为等比数列;
(Ⅱ)记{an}的公比为f(t),作数列{bn},使b1=1,,求和:。
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