The cost of the nth year is 12+4(n- 1)=8+4n (according to arithmetic progression's solution).

Then the expenditure in the first n years is 8n+2n(n+ 1)=2n(n+5) (according to arithmetic progression's solution).

Expenditure =2n(n+5)+98

Income =50n

When income > expenditure, it starts to make a profit, that is, 50n >；; 2n(n+5)+98

So in the third year, I began to make a profit.

(2) The average annual profit is 1 hour. =[50n-2n(n+5)-98]/n=40-2n-98/n, and the annual average profit is the largest, that is, n= 14, then the total profit of the company is 70+26 = 960,000.

2. When the total profit reaches the maximum value n= 10 (obtained by (1)), the company's * * * income 102+8 = 165438+ ten thousand.

Obviously, the second option is more cost-effective.

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