数列an满足a1=1 a2=3/2 an+2=3/2an+1-1/2an n属于正整数（n+2和n+1为角标） 令bn=3n-2求数列{an*bn}的前n

Question

数列an满足a1=1 a2=3/2 an+2=3/2an+1-1/2an n属于正整数（n+2和n+1为角标） 令bn=3n-2求数列{an*bn}的前n

Answer 1

因为an+2=3/2an+1-1/2an
所以an+2-an+1=1/2an+1-1/2an=1/2(an+1-an)
又因为a2-a1=3/2-1=1/2
所以,数列{an+1-an}为首项为1/2,公比为1/2的等比数列
即an+1-an=(1/2)^n
an+1-a1=(1/2)^n+(1/2)^(n-1)+……+1/2=1-(1/2)^n
所以an+1=2-(1/2)^n
an=2-(1/2)^(n-1)
an*bn=(3n-2)(2-(1/2)^(n-1))=3n-2-(3n-2)*(1/2)^(n-1))
Sn=(1+4+……+(3n-2))-[1*1+4*(1/2)+……+(3n-2)*(1/2)^(n-1)]
令Tn=[1*1+4*(1/2)+……+(3n-2)*(1/2)^(n-1)]
可知,
2Tn-Tn=[1*2+4*1+……+(3n-2)*(1/2)^(n-2)]-[1*1+4*(1/2)+……+(3n-2)*(1/2)^(n-1)]
=2+3[1+(1/2)+……+(1/2)^(n-2)]-(3n-2)*(1/2)^(n-1)
=8-3*(1/2)^(n-2)-(3n-2)*(1/2)^(n-1)
=8-(3n-8)*(1/2)^(n-1)
又因为,1+4+……+(3n-2)=n(3n-1)/2
所以,Sn=n(3n-1)/2-Tn=n(3n-1)/2-8+(3n-8)*(1/2)^(n-1)