Sequence reasoning, 22,36,40,56,68 ( )

Sequence reasoning, 22,36,40,56,68 ( )

Typing trouble, can't type x1x2x3, use letters
Let the sequence be ABCDEFG XYZ, then from C, C = a + B / 2, d = B + C / 2, and so on, the conclusion is 56 + 68 / 2 is 90

Sequence formula derivation
The following is the process of deriving a formula "a = a + R (1-p ^ n) / (1-p)"
A = P * a + Q (item n + 1 of table a)
a=p*a+q
A-A = P (A-A)
Let r = A-A
So A-A = P (A-A) = PR; A-A = P (A-A) = P ^ 2 * r
And so on
a=a+(a-a)+(a-a)+.+(a-a)+(a-a)
=A + R + PR + P ^ 2 * r +. + P ^ (n-1) * r
=a+r(1-p^n)/(1-p)
But a = a + (A-A) + (A-A) +. + (A-A) + (A-A) = a + R + PR + P ^ 2 * r +. + P ^ (n-1) * r
The following items after the second equal sign are p times of the preceding items, and (A-A) is (n-3) after (A-A). (A-A) corresponds to p ^ 1 * r, so (A-A) corresponds to p ^ (1 + n-3) * r = P ^ (n-2) * r
Can you help me find out the blind spot of this way of reasoning? Thank you very much

If your recursive formula a = P * a + Q is correct, the error is in the formula "a = a + R (1-p ^ n) / (1-p)". According to your recursive formula a = P * a + Q, the formula should be a = a + R (1-p ^) / (1-p). You can use the following test: a = P * a + Q, that is, A-Q / (1-p) = P * (A-Q / (1-p)), A-Q / (1-p) = P * (A-Q / (1-p)), {A-Q / (1-p)} is an equal ratio sequence with a common ratio of P (P is not equal to 1, as can be seen from the formula deduced), A-Q / (1-p) = (A-Q / (1-p)) * q ^ thus a = q / (1-p) + (A-Q / (1-p)) * q ^ = a = a + R (1-p ^) / (1-p), where r = A-A, which is completely consistent with your above derivation

Derivation of sequence formula
Recursion formula an = 2A (n-1) + 1, (n > 1), A1 = 1 (or A1 = a), to find the general term formula of an? It's better to add a method. High school students have long forgotten this general term formula. Now, if you want this general term formula, you can review it later. The sequence seems to be useful

An=2A(n-1)+1
An+1=2A(n-1)+2
An+1=2[A(n-1)+1]
{an + 1} is an equal ratio sequence with a common ratio of 2
A1 + 1 = 2 (if A1 = 1)
An+1=2^n
An=2^n-1
If A1 = a
Then an + 1 = [2 ^ (n-1)] * (a + 1)
An=[2^（n-1）]*(a+1)-1

How to deduce the fixed point method of sequence?

For example, the general form of implicit function theorem and inverse function theorem in mathematical analysis, and the existence and uniqueness theorem of solution of initial value problem of differential equation are all based on fixed point

May I ask the name of the derivation and proof method of the sum formula of the first n terms of the equal ratio sequence

See mathematics senior one compulsory 5 for details of dislocation subtraction method