# 3 questions.

1. May 26, 2004

### mustang

Problem 16.
Show that 1+2+4+...+26(n-1)=2^(n)-1
I first decided to use the formula: a_n=a_1*r^(n-1)

Problem24. Find the sum of the series 1-3+5-7+9-11+....+1001.

Problem 26. Suppose a doctor earns \$40,000 during the first year of practice. Suppose also that each succeeding year the salary increases 10%. What is the total of the doctor's salaries over the first 10 years? How many years must the doctor work if the salary total is to exceed a million dollars?

2. May 26, 2004

### AKG

That's not the correct formula. If you want to find the nth term in a sequence that might be useful, but you want to find the nth term in a series. There is a different formula, I don't remember off-hand, I'm sure you can look it up or find it in your text book. However, if you need to prove it, you can do a simple proof by induction.

Notice that every consecutive pair of terms sums to -2, e.g. 1-2 = -2, 5-7 = -2, etc. Your answer should be 1001 - 2x, where x is the number of pairs. The first pair is (1,-3) and the last pair would be (997,-999). It's a simple matter of some simple division to find out x.

You can express this as a series. $a_1 = \ 40,000$, $r = 1.1$ because he gets a 10% increase each year, so his salary will be 10% greater than the previous year, which is 110% times more than the previous year, which is 1.1 times more than the previous year.