The superposition method, as the name implies, is shaped like

A[n]-a[n- 1] =f(n) find a [n]

Where f(n) is a typical arithmetic or proportional sequence, or it is relatively easy to find S[n].

for instance ...

A [n] = 4n+a [n- 1], a 1 = 1, find the general term.

therefore

a[n]-a[n- 1]=4n

a[n- 1]-a[n-2]= 4(n- 1)

a[n-2]-a[n-3]=4(n-2)

......

a[2]-a[ 1]=8

Add all the above n- 1 expressions (note that it is n- 1 expression, not n! )

a[n]-a[ 1]= 4n+4[n- 1]+...+8

a[n]- 1 = 8(n- 1)+(n- 1)(n-2)* 4/2

a[n]=8n-8 + 2(n？ -3n +2)+ 1

a[n]=2n？ +2n -3

PS 1: There are other ways to find a[n] in this question, because it is a formula in the form of a[n]=ka[n- 1]+f(n), and the answer can be obtained through construction.

PS2: If a[n]/a[n- 1]=f(n), the accumulation method can be obtained by analogy! I won't explain it here.