It is known that [an] and {BN} are arithmetic sequences, and CN = 3an + 2. It is proved that {CN} is also arithmetic sequences (2) The second question CN = an + BN (AB is not equal to 0): the sequence {CN} is an arithmetic sequence

It is known that [an] and {BN} are arithmetic sequences, and CN = 3an + 2. It is proved that {CN} is also arithmetic sequences (2) The second question CN = an + BN (AB is not equal to 0): the sequence {CN} is an arithmetic sequence

(1) Because CN = 3an + 2, C (n-1) = 3A (n-1) + 2,
Then: CN - C (n-1) = 3 [an-a (n-1)] = 3D (where D is the tolerance of the arithmetic sequence an)
Therefore, {CN} is an arithmetic sequence with C1 = (3A1 + 2) as the first term and 3D as the tolerance
(2) Because CN = an + BN, C (n-1) = a (n-1) + B (n-1),
Then: cn-c (n-1) = (an + BN) - [a (n-1) + B (n-1)]
=[An-A（n-1）]+[Bn-B（n-1）]
=D1 + D2 (where D1 and D2 are the tolerance of arithmetic sequence an and BN respectively)
Therefore, {CN} is an arithmetic sequence with C1 = (a1 + B1) as the first term and (D1 + D2) as the tolerance
Because an is an arithmetic sequence, let an = k * n + L (k, l are real numbers, n = 1, 2, 3.)
cn=3an+2=3*（k*n+l）+2=3k*n+（3l+2）；
C (n + 1) - C (n) = 3k, 3K is a real number, then there is a tolerance that CN is an arithmetic sequence
Solving linear equations: 2x1 + 3x3 = 1 x1-x2 + 2x3 = 1 x1-3x2 + 4x3 = 2
Solution: augmented matrix=
2 0 3 1
1 -1 2 1
1 -3 4 2
r1-2r2,r3-r2
0 2 -1 -1
1 -1 2 1
0 -2 2 1
r1+r3, r3*(-1/2), r2+r3
0 0 1 0
1 0 1 1/2
0 1 -1 -1/2
r2-r1,r3+r1
0 0 1 0
1 0 0 1/2
0 1 0 -1/2
Exchange bank
1 0 0 1/2
0 1 0 -1/2
0 0 1 0
The solutions of the equations are: (1 / 2, - 1 / 2,0) '
It is known that: as shown in the figure, in △ ABC, ab = AC, D is the midpoint of BC, de ⊥ AB, DF ⊥ AC, e and F are perpendicular feet, respectively
It is proved that: ∵ AB = AC, ∵ B = ∠ C, ∵ de ⊥ AB, DF ⊥ AC, ∵ bed = ∠ CFD = 90 °, ∵ D is the midpoint of BC, ∵ BD = CD, ≌ BDE ≌ △ CDF, ∵ be = CF,