# Find formula for series

1. Jul 23, 2014

### Maxo

1. The problem statement, all variables and given/known data
By using the general equation for an arithmetic series, find a formula for calculating the series
$$1+3+5+...+(2n-1)$$

2. Relevant equations
General equation for an arithmetic series:
$$\sum ^{n}_{k=1}k=\frac{n(n+1)}{2}$$

3. The attempt at a solution
Using the general equation we have
$$\sum ^{n}_{k=1}(2k-1)$$
but how can we go from there? What is the next step?

I tried replacing n with (2k-1) in the general equation but that gives the wrong answer. So how can we do it?

Last edited: Jul 23, 2014
2. Jul 23, 2014

### Mentallic

$$\sum (a+b) = \sum a + \sum b$$

and

$$\sum (ab) = a\sum b$$
if a is a constant that is independent of the summation index (k for example).

So using these formulae, you should be able to solve your problem.

edit:

And keep in mind that

$$\sum_{k=1}^{n}1 = 1+1+1+...+1 = n$$

because you go from k=1 to n while summing 1 each time (hence n times).

3. Jul 23, 2014

### Maxo

Allright, so we have

$$\sum ^{n}_{k=1}(2k-1)=\sum ^{n}_{1}2k-\sum ^{n}_{1}1$$
The second term is easy since we see that we have 1 n times so 1*n = n as you already wrote. Can it be also calculated using the general equation? When I try I get (1(1+1))/2=2/2=1 which is not correct. Also if we just put in n it gets n(n+1)/2 which does not equal n! I guess I am using the general equation wrong? How should it be used then?

Then the first term, how is that calculated? Using the general equation I get 2k(2k+1)/2 = k(2k+1) which is again wrong. Please tell help me how to use the general equation.

4. Jul 23, 2014

### Mentallic

Notice the similarity?

$$\sum^n_1 2k =2\cdot 1+2\cdot 2+2\cdot 3+...+2\cdot n \\ = 2(1+2+3+...+n) \\ = 2\sum^n_1 k$$

Do you know how to prove the general summation formula? If you do, you'll also see why you can't use it in the instances that you've tried to. If not, begin with

$$S=1+2+3+...+n$$

and then consider

$$S=n+...+3+2+1$$

and add both of these values together, term by term (add the values in the same column in pairs)

$$2S=(1+n)+(2+(n-1))+(3+(n-2))+...+((n-2)+3)+((n-1)+2)+(n+1)$$

and simplify the right side, then solve for S.

Now, once you've learned how to prove the sum of the first n integers, now use a similar technique to find the general summation formula

$$S=a+(a+d)+(a+2d)+...+(a+(n-1)d)$$

where a is your starting point, and d is the difference between each value.

5. Jul 23, 2014

### Maxo

Thank you, you explain very well! I understand everything better now.
So if we write this again the opposite direction and then add column wise, we get that each term is equal to 2a+(n-1)d and we have n such terms, so 2S = n(2a+(n-1)d) = 2an + nd(n-1), so S = an + nd(n-1)/2.

In other words, the general formula for an arithmetic sum is:
$$Sn=an+\frac{n^2d-nd}{2}$$
Correct?
That's interesting, I have never seen this more "more general"(?) summation formula before. Does it have some particular name? In my math book, which is supposed to be for university level math, it's not even mentioned.

Btw if you have some suggestions where I can learn more and where it's explained as well as you explain here, please feel free to share it.

Last edited: Jul 23, 2014
6. Jul 23, 2014

### Mentallic

You're welcome! Yes, that's correct! Although, the factored form is more generally used.

$$S_n = \frac{n}{2}\left(2a+(n-1)d\right)$$

and now, if we let d=1 and a=1, then this says we are taking the sum of a linear series that starts with 1 (a=1) and increases by a value of 1 each time (d=1) which would give you your sum of positive integers. If you want to find

$$\sum_{k=1}^n 1$$

then this is like letting a=1 and d=0, which would give you S=n as expected.

Also,

$$\sum_{k=1}^n 2k = 2+4+6+...+2n$$

Which as we can see from the expanded form, we would let a=2 and d=2 since it starts at the value of 2 and increases by a value of 2 between each term. The formula, as well as the method I showed you earlier that moves the 2 outside of the sum would of course give you the same result of S=n(n+1)

It's called an arithmetic progression. Just google that and you'll find plenty on it

7. Jul 24, 2014

### Staff: Mentor

Isn't the formula for the sum of an arithmetic series n(a+l)/2, where a is the first term, l is the last term, and n is the number of terms?

Chet

8. Jul 25, 2014

### HallsofIvy

Yes, it is. Another way of looking at it is that the average value in an arithmetic sequence is the same as the average of the first and last term.

By the way, here's another way of summing that original series, $$\sum_{ix= 1}^n 2n-1$$. That is, of course, the sum of odd integers from 1 to 2n-1. It is the same as the sum of all integers from 1 to 2n minus the sum of all even numbers from 2 to 2n.

The sum of all numbers from 1 to 2n is, using that formula, $2n(1+ 2n)/2= n(2n+ 1)$. The sum of all even numbers is $2+ 4+ \cdot\cdot\cdot+ 2n= 2(1+ 2+ 3+ \cdot\cdot\cdot+ n)= 2(n(n+1)/2)= n(n+ 1)$.

9. Jul 31, 2014

### Maxo

Is there any way to write this general sum using the sigma notation? If so, how could it be written with Sigma notation?

10. Jul 31, 2014

### Mentallic

Well you aren't going to be writing the formula above using sigma notation, because sigma is used as a shorthand form to express summations that follow an expressible pattern. But the arithmetic sum itself can be expressed in sigma notation of course.

$$a+(a+d)+(a+2d)+...+(a+(n-1)d)$$
$$= \sum_{k=0}^{n-1}a+k\hspace{1 mm}d$$

This is a valid summation, but we can do more with it. Notice the constant a is added n times, hence to remove it from the sum, we exchange 'a' in the summation for 'an' outside of the summation. Also the constant d can be factored out of the sum. It's for the same reason that

$$1d+2d+3d+...+nd = d(1+2+3+...+n)$$

$$= an + d\sum_{k=0}^{n-1}k$$

Now at this point, k=0 clearly doesn't give us anything. We only needed k=0 in the first place for when a was within the sum (else if we started with k=1 then we'd only be adding the value of a (n-1) times.). So we can remove k=0 and start at k=1.

$$=an + d\sum_{k=1}^{n-1} k$$

And that would be your summation for a general arithmetic series.

11. Aug 1, 2014

### Maxo

Thank you, I would pretty much call this a perfect answer. :)