Sum of all integers from n to 2n

In summary, the problem is asking for the sum of all even numbers from 0 to 2n for any positive integer n. After some confusion about the wording, the correct understanding is that the even values from 0 to 4 should be included when n=2, resulting in a sum of 6. The formula (n^n)+n is proposed as a solution, but it is pointed out that this only works for adding all even numbers, whereas Gauss's formula is more general for adding all numbers. Suggestions for solving the problem are given, including considering the sum of all numbers from 1 to n and then multiplying by 2, and using a similar technique to Gauss's method.
  • #1
iamhumble
28
0

Homework Statement



What is the sum of all the even numbers from 0 to 2n, for any positive integer n?

Homework Equations



does not apply

The Attempt at a Solution



On any other day I would be able to solve the problem within seconds. However, my brain is fried beyond repair at the moment.

First, I want to know if I understand the question correctly. So, I have the equation "0 to 2n" and when "n = 2" the equation will now be "0 to 4". Now, when I sum up all the even integers do I also include 4 or not? The word "to" is messing it up. I know it is lame but I can't think straight now.

I believe that I do include the even values from 0 to 4 when "n=2". If this is the case, the sum will be 2 + 4 = 6.

Here are other sum values I have computated:

when n = 3, sum is 2 + 4 + 6 = 12
when n = 4, sum is 2 + 4 + 6 + 8 = 20
when n = 5, sum is 2 + 4 + 6 + 8 + 10 = 30​

I have be doing my best to develop an equation to satisfy the sum for any positive inter n and the only thing I could come up with is this.

given a positive n integer, take (n^n) + n to get the sum of all the even positive integer value​

Wait. I just figured out the equation!

All I am asking assistance for is if I have the correct understanding above. Any suggestions will be nice.
 
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  • #2
You got it correct, and you also posted almost the correct answer.
There's a famous story about how Gauss added up the numbers 1 to 100 very quickly, which leads to some nice counting techniques.
 
  • #3
iamhumble said:

Homework Statement



What is the sum of all the even numbers from 0 to 2n, for any positive integer n?

Homework Equations



does not apply

The Attempt at a Solution



On any other day I would be able to solve the problem within seconds. However, my brain is fried beyond repair at the moment.

First, I want to know if I understand the question correctly. So, I have the equation "0 to 2n" and when "n = 2" the equation will now be "0 to 4". Now, when I sum up all the even integers do I also include 4 or not? The word "to" is messing it up. I know it is lame but I can't think straight now.

I believe that I do include the even values from 0 to 4 when "n=2". If this is the case, the sum will be 2 + 4 = 6.

Here are other sum values I have computated:

when n = 3, sum is 2 + 4 + 6 = 12
when n = 4, sum is 2 + 4 + 6 + 8 = 20
when n = 5, sum is 2 + 4 + 6 + 8 + 10 = 30​

I have be doing my best to develop an equation to satisfy the sum for any positive inter n and the only thing I could come up with is this.

given a positive n integer, take (n^n) + n to get the sum of all the even positive integer value​

Wait. I just figured out the equation!

All I am asking assistance for is if I have the correct understanding above. Any suggestions will be nice.

Johann Carl Friedrich Gauss solved this problem when he was 7, at elementary school.
The terms of your series are:
0, 2, 4, ... 2n-4, 2n-2, 2n.
Gauss figured that:
0 + 2n = 2n
2 + (2n-2) = 2n
4 + (2n-4) = 2n
...
Since there are n+1 terms from 0 to 2n, there are (n+1)/2 sums, so
[tex]Sum = \frac{2n\left(n+1\right)}{2}[/tex]
 
Last edited:
  • #4
Thanks for the feedback and the "counting techniques" web site. It appears very useful.

I believe I am correct with my equation of "(n^n)+n" because the question that was asked has the following to say..

"What is the sum of all the even numbers ..."

The keyword here is "even".

I do agree that if I have to add up all "even" and "odd" numbers then Guass formula could be used. Boy, do I wish to have his brain power.
 
  • #5
Aww come on dude, I'm sure no matter how brain fried you are you can see how the sum of the even numbers can be reduced to the same problem gauss solved? Try taking out a common multiple.
 
  • #6
iamhumble said:
Thanks for the feedback and the "counting techniques" web site. It appears very useful.

I believe I am correct with my equation of "(n^n)+n" because the question that was asked has the following to say..

"What is the sum of all the even numbers ..."

The keyword here is "even".

I do agree that if I have to add up all "even" and "odd" numbers then Guass formula could be used. Boy, do I wish to have his brain power.

The example I gave in my previous post considers only even numbers and I used Gauss's reasoning to derive it.
 
  • #7
Have you considered adding all the numbers from 1 to n and then multiplying by 2?
 
  • #8
iamhumble said:
I believe I am correct with my equation of "(n^n)+n" [...]
Perhaps then there is a miscommunication about the meaning of the symbol ^.
I read it as "to the power" and then for n = 2, 3 and 4 then I get
(2^2) + 2 = 4 + 2 = 6
(3^3) + 3 = 27 + 3 = 30
(4^4) + 4 = 256 + 4 = 260
respectively (disclaiming any errors in elementary multiplication :smile:).

Anyway, I'm sure you saw the pattern but it's also important to write it down correctly :wink:

As for your problem, suppose I have
0 2 4 6 8 10 12 14
Try applying the same technique Gauss (supposedly, I don't really trust the story) did, adding them from the outside inwards.
 

1. What is the formula for finding the sum of all integers from n to 2n?

The formula for finding the sum of all integers from n to 2n is (3n^2 + 3n)/2.

2. How do you prove the formula for finding the sum of all integers from n to 2n?

The formula can be proven using mathematical induction. First, the formula can be shown to hold for n = 1. Then, assuming the formula holds for n = k, it can be shown that it also holds for n = k+1. This proves that the formula holds for all positive integers n.

3. Can the sum of all integers from n to 2n be calculated without using the formula?

Yes, the sum can also be calculated by adding each integer individually from n to 2n. For example, if n = 1, the sum would be 1 + 2 + 3 + ... + 2n = (2n)(2n+1)/2. This method is known as the "naive method" and is less efficient for larger values of n.

4. What is the significance of the sum of all integers from n to 2n in mathematics?

The sum of all integers from n to 2n has various applications in mathematics, such as in the study of sequences and series, number theory, and combinatorics. It is also helpful in solving problems involving arithmetic and geometric progressions.

5. Are there any real-world applications of the sum of all integers from n to 2n?

Yes, the sum of all integers from n to 2n can be applied in real-world scenarios, such as calculating the total distance traveled by a car during a trip with varying speeds or finding the total cost of items with increasing prices. It can also be useful in budgeting and financial planning.

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