At a job fair, two companies, A and B, set salary standards respectively: Company A promised to work for the first year.

(1) If this person has worked in Company A or Company B for N consecutive years, what is his monthly salary in the N-th year?

(2) The person intends to work continuously in a company for 65,438+00 years, and the application standard is only the total salary (excluding other factors). Which company should this person choose and why?

(3) What is the highest monthly salary (accurate to 1 yuan) when working in Company A than in Company B? And explain why.

Solution: (1) The monthly salary of this person in Company A and Company B in the nth year is an =1500+230× (n-1) (n ∈ n *) and BN = 2000 (1+5%, respectively.

(2) If the person has worked in Company A for 10 years, his total wage income is12 (A1+A2+A10) = 304,200 yuan;

If this person has worked in company B for 10 years, then his total wage income is12 (b1+B2+b10) ≈ 301869 (yuan).

Because the total income in company A is higher, this person should choose company A. 。

(3) The problem is equivalent to finding the maximum value of cn = an-bn =1270+230n-2000x1.05n-1(n ∈ n *).

When n≥2, cn-cn-1= 230-100×1.05n-2.

When cn-cn- 1 > 0, that is, 230-100×1.05n-2 > 0, 1.05n-2 < 2.3, n <19./kloc.

Therefore, when 2≤n≤ 19, cn-1< cn; When n≥20, CN ≤ CN- 1.

∴c 19 is the largest item in the series {cn}, and c19 = a19-b19 ≈ 827 (yuan), that is to say, the monthly salary of working in company A can be up to 827 yuan more than that in company B.