已知数列 {Cn } 满足 Cn = an bn 其中 {an } 等差,｛bn｝是等比数列,求｛Cn...
2个回答

设等比数列｛bn｝的公比是q,前n项和为Tn,

等差数列{an}的公差为d,前n项和为T'n

Sn=c1+c2+…+cn

=a1b1+a2b2+…+anbn

qSn= a1b2+…+a(n-1)bn+anb(n+1)

两式错位相减得

(1-q)Sn=a1b1+(a2-a1)b2+…+[an-a(n-1)]bn-anb(n+1)

=a1b1+(b2+…+bn)d-anb(n+1)

然后用等比数列的前n项和公式就可求出Sn

当q=1时,bn=b1,Tn=nb1

Sn=a1b1+a2b1+…+anb1=(a1+a2+…+an)b1=b1*T'n

当q≠1时,Sn=[a1b1+(b2+…+bn)d-anb(n+1)]/(1-q)=[a1b1+(Tn-b1)d-anb(n+1)]/(1-q)

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