已知数列C（n+1)=5Cn+n,C1=0,依次代入Cn能否得到通项公式：Cn=(5^n-4n-1)/16呢?请分析一下

c(n+1) = 5c(n) + n,
c(n+1) + (n+1)x + y = 5[c(n) + nx + y] = 5c(n) + 5nx + 5y,
c(n+1) = 5c(n) + 4nx + 4y - x.
x = 1/4,0 = 4y-x,y = x/4 = 1/16.
c(n+1) + (n+1)/4 + 1/16 = 5c(n) + n + (n+1)/4 + 1/16 = 5c(n) + 5n/4 + 5/16 = 5[c(n) + n/4 + 1/16],
{c(n) + n/4 + 1/16}是首项为c(1) + 1/4 + 1/16 = 5/16,公比为5的等比数列.
c(n) + n/4 + 1/16 = (5/16)5^(n-1) = 5^n/16
c(n) = [5^n - 4n - 1]/16.

C(n+1)+n/4+5/16=5[Cn+(n-1)/4+5/16],b(n+1)=5bn,b1=5/16,bn=5/16*5的n-1次方，cn=
(5^n-4n-1)/16