At a job fair, Company A and Company B made their own salary standards: Company A promised that the salary for the first year would be 1500 yuan, and every year thereafter.

Solution: (1) The monthly salary of this person in the N year of working in A is

a N = 1500+230×(N- 1)(N∈N *)，

What is the monthly salary for the nth year of working in B?

b N = 2000( 1+5%)N- 1(N∈N *)。

This person has been working in Company A continuously for 10 years, and his total wage income is

12 (a1+a 2+…+a10) = 304200 (yuan);

This person has worked in Company B for 10 years, and his total wage income is

12 (b1+B2+…+b10) ≈ 301869 (yuan).

Because the total income of Company A is relatively high,

Therefore, this person should choose Company A. 。

(2) From the meaning of the question, let C n = A n ﹣ B n =1270+230n ﹣ 2000×1.05n ﹣1(n ∈ n *),

When n≥2, cn ﹣ cn ﹣1= 230 ﹣100x1.05n ﹣ 2.

When cn ﹣ cn ﹣ 1 > 0, that is, 230 ﹣100×1.05n ﹣ 2 > 0.

1.05n-2 < 2.3, where n < 19. 1.

Therefore, when 2≤n≤ 19, cn-1< cn;

When n≥20, Cn ≤ Cn- 1.

∴c 19 is the largest term of the sequence {c n}.

C19 = a19-b19 ≈ 827 (yuan),

In other words, working in company A can earn up to 827 yuan more than working in company B. 。