Question

已知数列{an}中,a1=5,an=2a(n-1)+2^n-1(n∈N*,n≥2) 若数列{an+λ/2^n}为等差数列

Answer 1

n≥2时,

an=2a(n-1)+2ⁿ -1

等式两边同除以2ⁿ

an/2ⁿ=a(n-1)/2^(n-1) +1 -1/2ⁿ

an/2ⁿ -1/2ⁿ=a(n-1)/2^(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.

(an+λ)/2ⁿ=(an -1)/2ⁿ=[an+(-1)]/2ⁿ

λ=-1,