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

已知数列{an}中,a1=-1,a2=2,且an+1+an-1=2(an +1)(n≥2,n∈N
已知数列{an}中,
a1=-1,a2=2,且an+1+an-1=2(an +1)(n≥2,n∈N+).
求证：1)数列{an-an-1}为等差数列（2）求通项an
注,a的旁边是角标,比a小,如n+1等,电脑不知道怎么小写

人气：161 ℃　时间：2020-03-26 18:18:18

解答

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