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

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In summary: The formula for this is: a_n = a_1 * r^(n-1). You need to find the sum of the first 10 terms in this series, so you need to find a_10. You can use the formula for the nth term of a geometric sequence to find this. a_n = a_1*r^(n-1) = \$ 40,000 * (1.1)^(10-1) = \$ 40,000 * (1.1)^9 = \$ 40,000 * 2.3579472... = \$ 94,317.888... So a_10 is approximately \$ 94,317.89. To find the sum
  • #1
mustang
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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?
 
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  • #2
mustang said:
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. [itex]a_1 = \$ 40,000[/itex], [itex]r = 1.1[/itex] 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.
 
  • #3


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
 
  • #4


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
 

1. How can 1+2+4+...+2^(n-1) be proven to equal 2^(n)-1?

The proof for this equation is based on mathematical induction. It involves demonstrating that the equation holds true for a base case (n=1), and then showing that if it holds true for any value of n, it must also hold true for n+1.

2. What is the base case for proving 1+2+4+...+2^(n-1)=2^(n)-1?

The base case for this proof is n=1. When n=1, the equation becomes 1=2^1-1, which is true.

3. How is the inductive step used in proving 1+2+4+...+2^(n-1)=2^(n)-1?

The inductive step involves assuming that the equation holds true for a certain value of n, and then using that assumption to prove that it holds true for n+1. This is done by adding 2^n (or the next term in the sequence) to both sides of the equation and simplifying.

4. Can the proof for 1+2+4+...+2^(n-1)=2^(n)-1 be generalized for any value of n?

Yes, the proof for this equation can be generalized for any positive integer n. This is because the inductive step relies on the assumption that the equation holds true for any value of n, not just a specific value.

5. Are there any alternative methods for proving 1+2+4+...+2^(n-1)=2^(n)-1?

Yes, there are alternative methods for proving this equation, such as using a geometric series or mathematical identities. However, the proof using mathematical induction is the most common and accepted method for this particular equation.

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