Question

已知数列{an}中,a1=-1,a2=2,且an+1+an-1=2(an +1)(n≥2,n∈N

Answer 1

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

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

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

{a(n+1)-a(n)}是首项为a(2)-a(1)=3,公差为2的等差数列.

a(n+1)-a(n) = 3 + 2(n-1) = 2n + 1 = n(n+1)-(n-1)n + (n+1)-n,

a(n+1) - n(n+1) - (n+1) = a(n) - (n-1)n - n,

{a(n) - (n-1)n - n}是首项为a(1) - 0 - 1 = -2,的常数数列.

a(n) - (n-1)n - n = -2,

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