In the sequence {an}, two adjacent terms an and an + 1 are the corresponding quadratic equations Two adjacent terms in sequence {an} an.an +1 is two of the equation x ^ 2 + 3nx + BN = 0. If A1 = 2, try to find the value of B100

In the sequence {an}, two adjacent terms an and an + 1 are the corresponding quadratic equations Two adjacent terms in sequence {an} an.an +1 is two of the equation x ^ 2 + 3nx + BN = 0. If A1 = 2, try to find the value of B100

Firstly, according to the relation between heel and coefficient: an + an + 1 = - 3N (1); an * an + 1 = BN (2)
From (1), we can get: an
Substituting (2) for BN

Let the sum of the first n terms of the sequence an set be the square of Sn = 2 * n, the BN set be an equal ratio sequence, and A1 = B1, B2 (a2-a1) = B1
Term formula

an=Sn-S(n-1)
=2n^2 -2(n-1)^2
=2n^2-2n^2-2+4n
=4n-2
∴a1=4×1-2=b1
a2=4×2-2=6
∵b2(a2-a1)=b1
∴q=b2/b1=1/(a2-a1)=1/4
∴bn=2·（1/4）^（n-1）=8(1/4)^n

The known sequence {an} satisfies A1 = 5, A2 = 5, a (n + 1) = an + 6A (n-1) (n ≥ 2)... I can't understand the answer
(1) It is proved that {a (n + 1) + 2An} is an equal ratio sequence;
(2) Find the general term formula of sequence {an};
(1) From a (n + 1) = an + 6A (n-1), a (n + 1) + 2An = 3 (an + 2 (an-1)) (n ≥ 2)
　　　　　　∵a1＝5,a2＝5　　∴a2＋2a1＝15
So the sequence {a (n + 1) + 2An} is an equal ratio sequence with 15 as the first term and 3 as the common ratio 5 points
(2) from (1), a (n + 1) + 2An = 5 &; 3 ^ n
From the method of undetermined coefficient, we can get (a (n + 1) - 3 ^ (n + 1)) = - 2 (an-3 ^ n)
That is, an-3 ^ n = 2 (- 2) ^ (n-1)
So an = 3 ^ n + 2 (- 2) ^ (n-1) = 3 ^ n - (- 2) ^ n 10 points
How to get (a (n + 1) - 3 ^ (n + 1)) = - 2 (an-3 ^ n) from the undetermined coefficient method

A (n + 1) + 2An = 5 &; 3 ^ n can deduce a (n + 1) = 5 &; 3 ^ n-2an = (3 + 2) &; 3 ^ n-2an = 3 &; 3 ^ n + 2 &; 3 ^ n-2an. This step should be understood. From this, we can get a (n + 1) = 3 &; 3 ^ n + 2 &; 3 ^ n-2an = 3 ^ (n + 1) - 2 (an-3 ^ n)
A (n + 1) - 3 ^ (n + 1) = - 2 (an-3 ^ n), do you understand?

It is known that the sequence {an} satisfies A1 = 5, A2 = 5, a (n + 1) = an + 6A (n-1), (n ≥ 2, n belongs to positive integer), if the sequence {a (n + 1) + an} is equal ratio sequence
1. Find all the input values, and find the general term formula of the sequence {an};
2. Proof: when k is odd, 1 / AK + 1 / a (K + 1)

For the form of a [n + 1] = P * a [n] + Q * a [n-1]
We can list an auxiliary equation to simplify the two solutions of the sequence x2 = PX + Q to x = x1, x = x2
So the original sequence can be changed to a [n + 1] - X1 * a [n] = X2 (a [n] + X1 * a [n-1])
(1) For this problem: x2 = x + 6, solve X1 = - 2, X2 = 3, then a [n + 1] + 2 * a [n] = 3 (a [n] + 2 * a [n-1])
The sequence {a [n + 1] + 2A [n]} is an equal ratio sequence
So a [n + 1] + 2A [n] = 5 * 3 ^ n → a [n + 1] - 3 ^ (n + 1) = - 2 (a [n] - 3 ^ n)
It is easy to find a [n] = 3 ^ n + 2 * (- 2) ^ (n-1)
（2）
When n = 1, 1 / A1 + 1 / A2 = 2 / 52
1/ak+1/a(k+1)
=1/(3^k+2^k)+1/(3^(k+1)-2^(k+1))
=(4*3^n-2^n)/((3^n+2^n)*(3^(n+1)-2^(n+1)))
And 3 ^ (n + 1) - 2 ^ (n + 1) = (3-2) (3 ^ n + 3 ^ (n-1) * 2 + +3^p*2^(n-p)+…… +2^n)
>(2²+2²+…… +2 & sup2;) (n + 1 2 & sup2;)
>3(n+1)
therefore
1/ak+1/a(k+1)