请教一道数列题{An}首相为1,且8倍的第n+1项与第n项的乘积减去16倍的第n+1项再加上2倍的第n项再加上5=0（n>=2）,设Bn=1/(数列A的第n项减去1/2)(n>=2),1.求{Bn},2.求{An*Bn}的前n项和Sn.

Question

请教一道数列题
{An}首相为1,且8倍的第n+1项与第n项的乘积减去16倍的第n+1项再加上2倍的第n项再加上5=0（n>=2）,设Bn=1/(数列A的第n项减去1/2)(n>=2),1.求{Bn},2.求{An*Bn}的前n项和Sn .

Answer 1

∵数列{a[n]},8a[n+1]a[n]-16a[n+1]+2a[n]+5=0 (n≥2)
∴8a[n+1](a[n]-2)=-(2a[n]+5)
即：a[n+1]=(2a[n]+5)/(16-8a[n])
采用不动点法,设：x=(2x+5)/(16-8x)
8x^2-14x+5=0,即：(4x-5)(2x-1)=0
∴x=5/4 或者 x=1/2
∴(a[n+1]-5/4)/(a[n+1]-1/2)
={(2a[n]+5)/(16-8a[n])-5/4}/{(2a[n]+5)/(16-8a[n])-1/2}
={4(2a[n]+5)-5(16-8a[n])}/{4(2a[n]+5)-2(16-8a[n])}
=(8a[n]+20-80+40a[n])/(8a[n]+20-32+16a[n])
=(48a[n]-60)/(24a[n]-12)
=(4a[n]-5)/(2a[n]-1)
=2(a[n]-5/4)/(a[n]-1/2)
∵a[1]=1
∴(a[1]-5/4)/(a[1]-1/2)=-1/2
∴{(a[n]-5/4)/(a[n]-1/2)}是首项为-1/2,公比为2的等比数列
即：(a[n]-5/4)/(a[n]-1/2)=(-1/2)2^(n-1)=-2^(n-2)
a[n]-5/4=-(a[n]-1/2)2^(n-2)
a[n]+a[n]2^(n-2)=5/4+2^(n-3)
∴a[n]=[5/4+2^(n-3)]/[1+2^(n-2)]=[5+2^(n-1)]/(4+2^n)
∵b[n]=1/(a[n]-1/2) (n≥2)
∴b[n]
=1/{[5+2^(n-1)]/(4+2^n)-1/2}
=(4+2^n)/{[5+2^(n-1)]-[2+2^(n-1)]}
=(4+2^n)/3
(2)∵a[n]b[n]={[5+2^(n-1)]/(4+2^n)}{(4+2^n)/3}=[5+2^(n-1)]/3
∴{a[n]b[n]}的前n项和S[n]
=a[1]b[1]+a[2]b[2]+...+a[n]b[n]
=[5+2^0]/3+[5+2^1]/3+...+[5+2^(n-1)]/3
=(5n+2^n-1)/3