Respuesta :
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]=[tex]$10,500(1.06)^2[/tex].
Similarly we can check that in the year 1987 the worker earned [tex]$10,500(1.06)^3[/tex], which makes the pattern clear.
We can count that from the year 1985 to 1987 we had 2+1 salaries, so from 1985 to 2010 there are 2010-1985+1=26 salaries. This means that the total paid salaries are:
[tex]10,500+10,500(1.06)^1+10,500(1.06)^2+10,500(1.06)^3...10,500(1.06)^{26}[/tex].
Factorizing, we have
[tex]=10,500[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]=10,500\cdot[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}][/tex]
We recognize the sum as the geometric sum with first term 1 and common ratio 1.06, applying the formula
[tex]\sum_{i=1}^{n} a_i= a(\frac{1-r^n}{1-r}) [/tex] (where a is the first term and r is the common ratio) we have:
[tex]\sum_{i=1}^{26} a_i= 1(\frac{1-(1.06)^{26}}{1-1.06})= \frac{1-4.55}{-0.06}= 59.17 [/tex].
Finally, multiplying 10,500 by 59.17 we have 621.285 ($).
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 [tex](1.06)^{26}[/tex].
Answer: D
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]=[tex]$10,500(1.06)^2[/tex].
Similarly we can check that in the year 1987 the worker earned [tex]$10,500(1.06)^3[/tex], which makes the pattern clear.
We can count that from the year 1985 to 1987 we had 2+1 salaries, so from 1985 to 2010 there are 2010-1985+1=26 salaries. This means that the total paid salaries are:
[tex]10,500+10,500(1.06)^1+10,500(1.06)^2+10,500(1.06)^3...10,500(1.06)^{26}[/tex].
Factorizing, we have
[tex]=10,500[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]=10,500\cdot[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}][/tex]
We recognize the sum as the geometric sum with first term 1 and common ratio 1.06, applying the formula
[tex]\sum_{i=1}^{n} a_i= a(\frac{1-r^n}{1-r}) [/tex] (where a is the first term and r is the common ratio) we have:
[tex]\sum_{i=1}^{26} a_i= 1(\frac{1-(1.06)^{26}}{1-1.06})= \frac{1-4.55}{-0.06}= 59.17 [/tex].
Finally, multiplying 10,500 by 59.17 we have 621.285 ($).
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 [tex](1.06)^{26}[/tex].
Answer: D
