Question

在数列{an}中,a1=3,an=-an-1-2n+1(n≥2,且n属于N*) (1)证明：数列{an+n}是等比数列,

Answer 1

1.

an=-a(n-1)-2n+1

an+n=-a(n-1)-n+1=-[a(n-1)+(n-1)]

(an +n)/[a(n-1)+(n-1)]=-1,为定值.

a1 +1=3+1=4

数列{an +n}是以4为首项,-1为公比的等比数列.

2.

an +n=4×(-1)^n=-4×(-1)ⁿ

an=-n -4×(-1)ⁿ

n=1时,a1=-1-4×(-1)=3,同样满足.

数列{an}的通项公式为an=-n-4×(-1)ⁿ