What is the formula for finding the sum of the first 50 odd numbers?

[gabrielh]

I'm working on a homework sheet that will represent where I stand mathematically in order to qualify for a summer PreCal course. I remember my Algebra 2 teacher showing me how to do this problem using a calculator, but I never learned how to do this by hand. Any help is appreciated.

Homework Statement

Find the sum of the first 50 odd numbers:

50
$$\sum\$$ 2n-1
n=1

Homework Equations

Summation problem.

The Attempt at a Solution

As I said, I've only solved this specific type of summation problem using a calculator, but I know how to do regular summation problems, for instance:

6
$$\sum\$$ 2n-1
n=1

2(1)-1 = 1
2(2)-1 = 3
2(3)-1 = 5
2(4)-1 = 7
2(5)-1 = 9
2(6)-1 = 11
-----
36

I hope that qualifies as an attempt.

 

[Cyosis]

Just doing the summation manually is kind of tedious. Do you know a formula for the sum of the first n natural numbers?

 

[gabrielh]

No I don't know the formula, but I'm sure it would help :)

 

[Cyosis]

Have you ever heard the story about Gauss adding up the first 100 natural numbers in mere seconds? I hope you have or this formula won't make much sense to you.

$$\sum_{k=1}^n k=\frac{n}{2}(n+1)$$

Use this sum to calculate the sum you're interested in.

 

[gabrielh]

Ouch, no I haven't.

 

[jgens]

Perhaps to help you derive a formula for the sum, let S be the finite sum,

S = n₀ + (n₀ + d) + (n₀ + 2d) + . . . + (n₀ + (n-1)d)

Where n₀ represents the first term of the sum and n represents the number of terms.

Since addition is communicative and associative, if we sum S twice we find that,

2S = [(n₀) + (n₀ + (n-1)d)] + [(n₀ + d) + (n₀ + (n-2)d)] + . . .

2S = (2n₀ + (n-1)d) + (2n₀ + (n-1)d) + . . .

Try to finish the derivation and see how it applies to the sum you're trying to compute. Sorry if this is hard to understand.

 

[jgens]

Cyosis, why would he/she use the sum for the first n natural numbers when his sum asks for odd natural numbers only?

 

[Cyosis]

If you know the sum for the first n natural numbers you can quickly solve the sum for the first odd numbers. Had he known it, it would have been the easiest way to calculate the sum he's interested in.

 

[Manchot]

Ouch, no I haven't.

The story is that a six-year-old Gauss was told by his teacher to add the numbers 1 to 100 as a form of busywork. In seconds, he came up with the answer, much to his teacher's surprise. What he did was notice that since 1+100=101, 2+99=101, 3+98=101, etc., the sum becomes a simple multiplication: there are 50 of the above "pairs," so the sum is just 50*101 = 5050. Anyway, that's the intuition behind the identity that Cyosis posted. You can pretty much use the exact same intuition.

 

[gabrielh]

The story is that a six-year-old Gauss was told by his teacher to add the numbers 1 to 100 as a form of busywork. In seconds, he came up with the answer, much to his teacher's surprise. What he did was notice that since 1+100=101, 2+99=101, 3+98=101, etc., the sum becomes a simple multiplication: there are 50 of the above "pairs," so the sum is just 50*101 = 5050. Anyway, that's the intuition behind the identity that Cyosis posted. You can pretty much use the exact same intuition.

Okay thanks, I get it now.

 

[Cyosis]

Are you able to use this information so far? Depending on where you are in your education we may want to use different methods. Is this a first course on summations, do you know what arithmetic progression is?

 

[Manchot]

Also, one of the things I like about Wolfram Alpha is that it tries to guess what series you're trying to input, as long as it's "simple." So, if you type in 1+3+5+..., it guesses that you're entering a sum of odd numbers, and spits out an analytical expression for the partial sum.

 

[gabrielh]

Are you able to use this information so far? Depending on where you are in your education we may want to use different methods. Is this a first course on summations, do you know what arithmetic progression is?

Well, I get some of it, but I'm thinking there may be an easier way for me. I just came out of an algebra 2 course where we learned simple summations like the one I used as an example in my original post. We only did one or two of the type of summation I'm currently asking about, and it was all on the calculator with no formula, we used pre-programed apps.

Is an arithmetic progression the same as an arithmetic sequence? I apologize if that's a stupid question.

 

[jgens]

Well, if you're interested, here's a couple of derivations for summation formulas. Sometimes knowing how formulas are derived is useful (at least I find it useful).

Arithmetic Sums:

Suppose we have an arithmetic sequence with a constant difference d between terms. We let the sum, S_A represent the sum of the first n terms of that sequence such that

S_A= k₀ + k₁ + . . . + k_n-1 + k_n

Where k₀ represents the first term of the series, k₁ = k₀ + d, k₂ = k₁ + d, and so on and so forth. Given our definition of k_i it follows that,

S_A = k₀ + (k₀ + d) + . . . + (k₀ + (n-2)d) + (k₀ + (n-1)d)

Since addition is both communicative and associative, we can take the sum twice and rearrange the terms such that,

2S_A = (k₀ + k_n) + (k₁ + k_n-1) + . . .

Therefore,

2S_A = [k₀ +(k₀ + (n-1)d)] + [(k₀ + d) + (k₀ + (n-2)d)] + . . .

2S_A = (2k₀ + (n-1)d) + (2k₀ + (n-1)d) + . . .

Since we have n number of terms in the sum, it follows that,

2S_A = n(2k₀ + (n-1)d)

and consequently, we have that,

S_A = (n/2)(2k₀ + (n-1)d)


From this, we see that the sum of the first n natural numbers is just a specific case of an arithmetic sequence. We know that the first term, k₀ = 1 and we know that d = 1. So the sum s_A of the first n natural numbers is,

s_A = (n/2)(n+1)


Geometric Sums:

Suppose we have a geometric series with a constant rate r between terms. We let the sum S_G represent the sum of the first n terms of that series such that,

S_G = k₀ + k₁ + . . . + k_n-1 + k_n

Where k₁ = k₀(r), k₂ = k₁(r), and so on and so forth. Ultimately, using these definitions, it follows that,

S_G = k₀ + k₀(r) + k₀(r)² + . . . + k₀(r)^n-1

If we multiply this sum by the constant rate r we find that,

(r)S_G = k₀(r) + k₀(r)² + k₀(r)³ + . . . + k₀(r)ⁿ

Finding the difference between the sums yields (the algebra is up to you),

S_G - (r)S_G = k₀ - k₀(r)ⁿ

Therefore,

S_G(1 - r) = k₀(1 - rⁿ)

and consequently,

S_G = k₀(1 - rⁿ)/(1 - r)


Hopefully you'll find this interesting or useful. If not, oh well.

 

[HallsofIvy]

1+ 3+ 5+ 7+ ...+ 93+ 95+ 97+ 99
99+ 97+ 95+ 93+...+ 7+ 5+ 3+ 1

What is the sum of each vertical column? How many columns are there? Notice that in adding each column and then getting the total you are adding "first 50 odd numbers" twice.

Another way:
1+ 3+ 5+ ...+ 45+ 47+ 49
99+ 97+ 95+ ...+ 55+ 53+ 51

What is the sum of each column? Now how many columns are there? So what is the sum of all the numbers. Notice that here, you have added each number exactly once.

 

[Дьявол]

First, if you don't know how many numbers are there (the number n), find it using:
$$a_n=a_1+(n-1)*d$$

where $a_n$ is the last number 49, $a_1$ is the first number 1, and $d$ is the difference between two numbers which equals 2.

Just substitute and find n.

Then, use the formula for finding the sum of n numbers of arithmetic progression:

$$S_n=\frac{n}{2}*[a_1+a_n]$$

Good luck.

Regards.

 

[gabrielh]

Well, if you're interested, here's a couple of derivations for summation formulas. Sometimes knowing how formulas are derived is useful (at least I find it useful).

Arithmetic Sums:

Suppose we have an arithmetic sequence with a constant difference d between terms. We let the sum, S_A represent the sum of the first n terms of that sequence such that

S_A= k₀ + k₁ + . . . + k_n-1 + k_n

Where k₀ represents the first term of the series, k₁ = k₀ + d, k₂ = k₁ + d, and so on and so forth. Given our definition of k_i it follows that,

S_A = k₀ + (k₀ + d) + . . . + (k₀ + (n-2)d) + (k₀ + (n-1)d)

Since addition is both communicative and associative, we can take the sum twice and rearrange the terms such that,

2S_A = (k₀ + k_n) + (k₁ + k_n-1) + . . .

Therefore,

2S_A = [k₀ +(k₀ + (n-1)d)] + [(k₀ + d) + (k₀ + (n-2)d)] + . . .

2S_A = (2k₀ + (n-1)d) + (2k₀ + (n-1)d) + . . .

Since we have n number of terms in the sum, it follows that,

2S_A = n(2k₀ + (n-1)d)

and consequently, we have that,

S_A = (n/2)(2k₀ + (n-1)d)


From this, we see that the sum of the first n natural numbers is just a specific case of an arithmetic sequence. We know that the first term, k₀ = 1 and we know that d = 1. So the sum s_A of the first n natural numbers is,

s_A = (n/2)(n+1)


Geometric Sums:

Suppose we have a geometric series with a constant rate r between terms. We let the sum S_G represent the sum of the first n terms of that series such that,

S_G = k₀ + k₁ + . . . + k_n-1 + k_n

Where k₁ = k₀(r), k₂ = k₁(r), and so on and so forth. Ultimately, using these definitions, it follows that,

S_G = k₀ + k₀(r) + k₀(r)² + . . . + k₀(r)^n-1

If we multiply this sum by the constant rate r we find that,

(r)S_G = k₀(r) + k₀(r)² + k₀(r)³ + . . . + k₀(r)ⁿ

Finding the difference between the sums yields (the algebra is up to you),

S_G - (r)S_G = k₀ - k₀(r)ⁿ

Therefore,

S_G(1 - r) = k₀(1 - rⁿ)

and consequently,

S_G = k₀(1 - rⁿ)/(1 - r)


Hopefully you'll find this interesting or useful. If not, oh well.

Thank you, this helps a lot.

 

[gabrielh]

First, if you don't know how many numbers are there (the number n), find it using:
$$a_n=a_1+(n-1)*d$$

where $a_n$ is the last number 49, $a_1$ is the first number 1, and $d$ is the difference between two numbers which equals 2.

Just substitute and find n.

Then, use the formula for finding the sum of n numbers of arithmetic progression:

$$S_n=\frac{n}{2}*[a_1+a_n]$$

Good luck.

Regards.

Thanks, I think I got it now.