Question

关于数列的问题已知数列{an}满足an=2an-1+2的n次方-1（n属于正自然数,n大于等于2）,且a4=81(1)求

Answer 1

1、

a4=2a3+2^4 -1

a3=(a4 -2^4 +1)/2=(81-16+1)/2=33

a3=2a2+2^3 -1

a2=(a3-2^3 +1)/2=(33-8+1)/2=13

a2=2a1+2^2 -1

a1=(a2-2^2 +1)/2=(13-4+1)/2=5

a1=5 a2=13 a3=33

2、

an=2a(n-1) +2ⁿ -1

an -1=2a(n-1) -2 +2ⁿ

等式两边同除以2ⁿ

(an -1)/2ⁿ=[a(n-1) -1]/2^(n-1) +1

(an -1)/2ⁿ-[a(n-1) -1]/2^(n-1)=1,为定值.

(a1-1)/2=(5-1)/2=2

数列{(an -1)/2ⁿ}是以2为首项,1为公差的等差数列.

λ=-1

(an -1)/2ⁿ=2+(n-1)=n+1

an=(n+1)×2ⁿ +1

数列{an}的通项公式为an=(n+1)×2ⁿ +1