Proving 1+2+4+...+2^(n-1)=2^(n)-1

[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?

 

[AKG]

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)

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 textbook. However, if you need to prove it, you can do a simple proof by induction.

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

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.

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?

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.

 

[M98Ranger]

To prove that 1+2+4+...+2^(n-1)=2^(n)-1, we can use mathematical induction. First, we will prove the base case, n=1.

When n=1, the left side of the equation becomes 1, and the right side becomes 2^1-1=1. Therefore, the equation holds true for n=1.

Next, we will assume that the equation holds true for some arbitrary value k, where k is a positive integer.

This means that 1+2+4+...+2^(k-1)=2^k-1.

Now, we will prove that the equation also holds true for n=k+1.

When n=k+1, the left side of the equation becomes 1+2+4+...+2^(k-1)+2^k.

By our assumption, we know that 1+2+4+...+2^(k-1)=2^k-1.

Substituting this into the left side of the equation, we get 2^k-1+2^k.

Using the properties of exponents, we can rewrite this as 2^k+2^k=2^(k+1).

Therefore, the left side of the equation becomes 2^(k+1), which is equal to the right side of the equation.

Thus, we have proven that 1+2+4+...+2^(n-1)=2^(n)-1 for all positive integer values of n.

Moving on to Problem 16, we can use the same formula a_n=a_1*r^(n-1) to find the sum of the series.

In this case, a_1=1 and r=2, since each term is twice the previous one.

So, a_n=1*2^(n-1)=2^(n-1).

To find the sum of the series, we can use the formula for the sum of a geometric series:

S_n=a_1*(1-r^n)/(1-r)

Plugging in our values, we get S_n=1*(1-2^n)/(1-2)=1*(1-2^n)/(-1)=2^n-1.

Therefore, the sum of the series 1+2+4+...+2

 

[M98Ranger]

To prove this statement, we can use mathematical induction.

First, we will show that the statement is true for n=1.
1+2^(1-1)=1+1=2
2^1-1=2-1=1
Thus, the statement is true for n=1.

Next, we will assume that the statement is true for some positive integer k.
1+2+4+...+2^(k-1)=2^k-1

We will now prove that the statement is also true for n=k+1.
1+2+4+...+2^(k-1)+2^k=2^(k+1)-1
Using our assumption, we can rewrite the left side as:
2^k-1+2^k=2^(k+1)-1
Therefore, the statement is true for n=k+1.

Since we have shown that the statement is true for n=1 and that it being true for n=k implies it is true for n=k+1, we can conclude that the statement is true for all positive integers n. Hence, 1+2+4+...+2^(n-1)=2^(n)-1 is true.

Problem 24.
To find the sum of the given series, we can group the terms in pairs.
1-3= -2
5-7= -2
9-11= -2
...
1001-x= -2
We can see that there are 500 pairs of -2, so the sum of the series is -2*500=-1000.

Problem 26.
The doctor's salary over the first 10 years can be represented as:
$40,000 + $44,000 + $48,400 + ... + $1,048,576
This is a geometric series with a common ratio of 10%. Using the formula for the sum of a geometric series, we can calculate the total salary over 10 years to be $2,097,151.

To find the number of years it takes for the total salary to exceed a million dollars, we can set up the following inequality:
$40,000 + $44,000 + $48,400 + ... + $1,048,576 > $1,000,000
Using the formula for the sum of a geometric