Question

求极限🆘求大佬

第119题求极限，答案里的an怎么解出来的呢

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Answer 1

a1=1, a2=2
3a(n+2)-4a(n+1)+an =0
3a(n+2) - a(n+1) = 3a(n+1) -an
=> { 3a(n+1) -an } 是等差数列, d=0
3a(n+1) -an = 3a2 -a1
= 6-1
=5
3a(n+1) -an = 5
a(n+1) = (1/3)an +5/3
a(n+1) -5/2 = (1/3)( an - 5/2)

=> {an -5/2} 是等比数列, q=1/3

an -5/2 =(1/3)^(n-1) . ( a1 -5/2)
=-(3/2). (1/3)^(n-1)
an =5/2 -(3/2). (1/3)^(n-1)
lim(n->无穷) an = 5/2