a[n+1] = 4a[n] - 3n + 1 = 4a[n] - 4n + n + 1
因此a[n+1] - (n+1) = 4a[n] - 4n
即b[n+1] = 4b[n],也就是说b[n]是等比数列
又b[1] = a[1] - 1 = 1
所以b[n] = 4^(n-1),所以其前n项和为(4^n - 1)/3
因此S[n] = (4^n-1)/3+n(n+1)/2