Given A1 = 1, an + 1 = an + 2n, finding an is known from the recurrence formula: a2-a1 = 2, a3-a2 = 22, a4-a3 = 23 An-an-1 = 2N-1 add the above n-1 to get an = a1 + 2 + 2x2 + 2x3 + 2x4 + +2(n-1)=1+2+2X2+2X3+… +2(n-1)=2(n-1) Why an = a1 + 2 + 2x2 + 2x3 + 2x4 + +2(n-1)=1+2+2X2+2X3+… +2 (n-1) = 2 (n-1)? Just like a1 + A2 + a3 + A4 +. An is equal to an in the end? (when D and A1 are all greater than zero!) And I want to know the specific steps and process of formula addition

Given A1 = 1, an + 1 = an + 2n, finding an is known from the recurrence formula: a2-a1 = 2, a3-a2 = 22, a4-a3 = 23 An-an-1 = 2N-1 add the above n-1 to get an = a1 + 2 + 2x2 + 2x3 + 2x4 + +2(n-1)=1+2+2X2+2X3+… +2(n-1)=2(n-1) Why an = a1 + 2 + 2x2 + 2x3 + 2x4 + +2(n-1)=1+2+2X2+2X3+… +2 (n-1) = 2 (n-1)? Just like a1 + A2 + a3 + A4 +. An is equal to an in the end? (when D and A1 are all greater than zero!) And I want to know the specific steps and process of formula addition

A2, A3 and a (n-1) all have positive and negative terms, which offset each other

The process of calculating the general term of a sequence by adding up!
We have solved an by adding one = an + 4N Square-1 / 1 under A1 = half an = 1 in an

Tip: 4N square - 1 / 1 = 1 / 2 [(2n-1) 1 / 2 minus (2n + 1) 1 / 2] get: a (n + 1) - A (n) = 1 / 2 [(2n-1) 1 / 1 minus (2n + 1) 1 / 1] believe you will add it? Get a (n + 1) - A1 = 1 / 2 [1 - (2n + 1) 1 / 1] so a (n + 1) = 1 - (4N + 2) 1 / 1, so a (...)

How to solve the problem of accumulation in sequence?
In the sequence {an}, an / a (n-1) = n-1 / N + 1, and then multiply term by term, an = (n-1 / N + 1) * (n-2 / N) / (n-3 / n-1) * ··· * (2 / 4) / (1 / 3) / A1. Why is this step?

In the sequence {an}, an / a (n-1) = (n-1) / (n + 1), and then multiply term by term, an = [(n-1) / (n + 1)] * [(n-2) / N] * [(n-3) / (n-1)] * ···· * (2 / 4) * (1 / 3) * A1. This is because the nth term an of any sequence without zero term can be written as an = [an / a (n-1)] * [a (n-1) / a (n-2)]

An = 2 × an-1 + n ^ 2 + 3 use the undetermined coefficient method to find the general formula of an

You mean an = 2A (n-1) + n & sup2; + 3?
Since the highest power is 2, there are 3 undetermined numbers of degree 2, degree 1 and degree 0
Let an + xn & sup2; + yn + r = 2 [a (n-1) + X (n-1) & sup2; + y (n-1) + R]
After expansion, it is arranged as an = 2A (n-1) + xn & sup2; + n (y-4x) + r-2y + 2x
The comparison coefficient is: x = 1, y-4x = 0, r-2y + 2x = 3
That is, x = 1, y = 4, r = 9
So an + n & sup2; + 4N + 9 = 2 [a (n-1) + (n-1) & sup2; + 4 (n-1) + 9]
Obviously {an + n & sup2; + 4N + 9} is an equal ratio sequence with a common ratio of 2
If you don't give A1, you can't work it out. You must know A1
If A1 is given, then a1 + 13 is the first term of {an + n & sup2; + 4N + 9}
So an + n & sup2; + 4N + 9 = (a1 + 13) * 2 ^ (n-1)
So an = [(a1 + 13) * 2 ^ (n-1)] - N & sup2; - 4n-9