Question

已知数列{an}满足a1=4,a(n+1)an+6a(n+1)-4an-8=0,记bn=6/an-2,n属于N*(1)求

Answer 1

n=6/(an-2),an=(6/bn)+2 a(n+1)=(6/b(n+1))+2,b1=3

把an=(6/bn)+2 a(n+1)=(6/b(n+1))+2代入a(n+1)an+6a(n+1)-4an-8=0化简得

3+4bn=4b(n+1) 即4(bn+1)=b(n+1)+1

bn+1为b1+1为首,4为公比的等比数列

bn+1=(b1+1)*4^(n-1)

bn=4^n-1

an*bn=6+2bn=2*4^n+4

Sn=4n+2*4*(1-4^n)/(1-4)=4n+8/3*(4^n-1)