Question

a(n+1)=3a(n)+5×2^n+4,a1=1,求数列a(n)的通项公式.PS：括号中是下标,2^n是2的n次方

Answer 1

a(n+1)=3a(n)+5×2^n+4

a(n+1)/2^(n+1)=3/2 a(n)/2^n+5/2+1/2^(n-1)

a(n+1)/2^(n+1)+1/2^n+5=3/2( a(n)/2^n+1/2^(n-1)+5)

a(1)/2+1/2^0+5=13/2,a(n)/2^n+1/2^(n-1)+5=13/2(3/2)^(n-1)

最后得a(n)=13*3^(n-1)-5*2^(n)-2