已知数列{an}满足a1=1 ,an+1=3an+2的n+1次幂,求an

a(n+1)=3an+2^(n+1)设a(n+1)+k*2^(n+1)=3[a(n)+k*2^n]则 a(n+1)=3an+3k*2^n-2k*2^n=3an+k*2^n所以 k=2即 a(n+1)+2*2^(n+1)=3[a(n)+2*2^n]所以 {an+2*2^n}是等比数列首项为a1+2*2^1=5,公比为3所以 an+2*2^n=5*3^(n-1)...