Aworker was paid a salary of $10,500 in 1985. each year, a salary increase of 6% of the previous year's salary was awarded. how much did the worker earn from the beginning of 1985 through the end of 2010? a $45,064.64 b $256,527.63 c $576,077.38 d $621,142.02

Note that 6% converted to a decimal number is 6/100=0.06. Also note that 6% of a certain quantity x is 0.06x.

Here is how much the worker earned each year:

In the year 1985 the worker earned $10,500. 

In the year 1986 the worker earned $10,500 + 0.06($10,500). Factorizing $10,500, we can write this sum as:

                                            $10,500(1+0.06).

In the year 1987 the worker earned

$10,500(1+0.06) + 0.06[$10,500(1+0.06)].

Now we can factorize $10,500(1+0.06) and write the earnings as:

$10,500(1+0.06) [1+0.06]=.

Similarly we can check that in the year 1987 the worker earned , which makes the pattern clear. 

We can count that from the year 1985 to 1987 we had 2+1 salaries, so from 1985 to 2010 there are 2010-1985+1=26 salaries. This means that the total paid salaries are:

.

Factorizing, we have



We recognize the sum as the geometric sum with first term 1 and common ratio 1.06, applying the formula

(where a is the first term and r is the common ratio) we have:

.

Finally, multiplying 10,500 by 59.17 we have 621.285 ($).

The answer we found is very close to D. The difference can be explained by the accuracy of the values used in calculation, most important, in calculating .

D