a(n+1)= 3an + 2^n
a(n+1) +2^(n+1)= 3(an + 2^n)
{an + 2^n} 是等比數(shù)列,q=3
an + 2^n = 3^(n-1) .(a1 + 2^1)
an = -2^n + (a1+2).3^(n-1)第二行是怎么想到的？a(n+1)= 3an + 2^na(n+1) +k.2^(n+1)= 3(an + k.2^n)2^n的系數(shù)k = 1我自己想明白了，謝謝你！