Why do we need to verify n = 1 when we use the undetermined coefficient method to find the general term formula

Why do we need to verify n = 1 when we use the undetermined coefficient method to find the general term formula

Because when n = 1, this method is not applicable. This involves the knowledge of advanced mathematics, so it will not be explained in detail. If you want to know, you can refer to relevant mathematics books

Find the general term formula of recursive sequence a (n) = (n-1) a (n-1) + 1, n > 1; A1 = 1?

Formula 1: a (n) = (n-1) * a (n-1) + 1
Formula 2: a (n-1) = (n-2) * a (n-2) + 1, multiply both sides by (n-1) at the same time
(n-1) * a (n-1) = (n-1) (n-2) * a (n-2) + (n-1)
Formula 3: (n-1) (n-2) * a (n-3) = (n-1) (n-2) (n-3) * a (n-3) + (n-1) (n-2)
…………
Formula n (n-1) (n-2)... 2 * a (2) = (n-1) (n-2)... 1 * a (1) + (n-1) (n-2)... 2
Add up the above n formulas and omit the same terms on both sides of the formula
a(n) = (n-1)! * a(1) + 1 + (n-1) + (n-1)(n-2) + .+(n-1)(n-2)...2
Substitute a (1) = 1, and multiply the denominator of each term by an appropriate factor
= (n-1)!/0! + (n-1)!/(n-1)! + (n-1)!/(n-2)! + (n-1)!/(n-3)! + ... +(n-1)!/1!
Extract common factors
=(n-1)! * sum (1 / I!), where I = from 0 to n-1

6. Try to find the sequence a (n) satisfying the following recurrence formula by using the method of characteristic root equation
(1) a(n+2)=a(n+1)+2a(n) ,a(0)=1,a(1)=0;
(2) a(n+2)=3a(n+1)-2a(n),a(0)=0,a(1)=1
(3) a(n+3)-2a(n+2)-a(n+1)+2a(n)=0,a(0)=0,a(1)=0,a(2)=6
(1) a(n)=[(2^n)+2*(-1)^n]/3
(2) a(n)=(2^n)-1
(3) a(n)=2^(n+1)-3+(-1)^n
Please write the detailed process and thinking

(1) The characteristic equation of: a (n + 2) = a (n + 1) + 2A (n) is: x ^ 2 = x + 2, x = - 1,2;
Let a (n) = C1 * (- 1) ^ n + C2 * 2 ^ n, a (0) = 1, a (1) = 0,
c1+c2=1,
-C1 + 2c2 = 0
c1=2/3,c2=1/3,
a(n)=[(2^n)+2*(-1)^n]/3
(2) The method is the same as above
(3) The characteristic equation is x ^ 3-2x ^ 2-x + 2 = 0, (x-1) (X-2) (x + 1) = 0, and the characteristic root is 1,2, - 1
Let a (n) = C1 * 1 ^ n + C2 * (- 1) ^ n + C3 * 2 ^ n = C1 + C2 * (- 1) ^ n + C3 * 2 ^ n
A (0) = 0, a (1) = 0, a (2) = 6 are brought into the solution to get C1, C2, C3
The general characteristic root equation method is used to find the general term. First, write out its characteristic equation, find out the characteristic root (x1, X2, X3..., XM), and then let the general term be: a (n) = C1 * X1 ^ n + C2 * x2 ^ n +... + cm * XM ^ n
According to the first m-Term solution of the equations, C1, C2,..., CM are obtained