Coming up with a formula for

1. Sep 12, 2004

Caldus

Um, how can I figure out a formula for solving something like:

1 + 3 + 5 + 7 + 9 + ... + n

In particular: 1 + 3 + 5 + ... + 999

Thanks.

2. Sep 12, 2004

geometer

As far as I know, there is no rote procedure for doing this. It's pretty much a trial and error process. Make yourself a table and figure out the pattern. In this case, we can write this as the sum from x = 0 to x = (n-1)/2 of 2x+1:

x = 0 ===> Sum = 1
x = 1 ===> Sum = 4
x = 2 ===> Sum = 9
x = 3 ===> Sum = 16

We can keep this up if we need to to help us see the pattern, but in this case, a little thought shows that the formula is: Sum from x= 0 to x = n = (n+1)^2

3. Sep 12, 2004

Integral

Staff Emeritus
I believe that Gauss did this as school child when given the assignment to sum the digits 1-100.

First note that if you add them together smallest to largest you get a constant result ( 100+ 1=101, 99+2=101 ..) multiply this by 100 because there are 100 sums and divide by 2 because you added all the numbers twice.

edit:
OPPS! just noticed that you were not summing ALL integers.

Last edited: Sep 12, 2004
4. Sep 12, 2004

Integral

Staff Emeritus
After a bit more playing,

consider this

$$S = \Sigma _ {i=1}^N (2i -1)$$

now using the properties of the summation

$$S= 2\Sigma_{i=1}^N i - \Sigma_{i=1}^N 1$$

$$S= 2\Sigma_{i=1}^N i -N$$

Now using Gauss's trick we can see that the sum of n integers is given by

$$S_n = \frac {(n+1)(n)} 2$$

So we can write for the sum of the odd integers

$$S = (N+1)(N)-N = N^2$$

Note that I have given the general expression for the odd integers as 2n-1 and my sum starts with n=1 where geometer used 2n+1 and started with n=0. This is the reason he has N+1 where I have N. Also note that this N is NOT the odd integer you are summing to but the NUMBER OF TERMS in the sum.

Last edited: Sep 12, 2004
5. Sep 12, 2004

shmoe

Integral, the same idea of pairing highest and lowest will work just fine,
1+3+5+7+9+11=12+12+12=(6/2)*12, where 6 is the number of terms here.

If you already have a formula for the sum of the first n integers, 1+2+3+..+n, you can use this as well:
1+3+5+7+9+11=(1+2+..+11)-(2+4+6+8+10)=(1+2+..+11)-2(1+2+3+4+5)=(insert formula here)-2(insert formula here)

Caldus, do you have a formula for the sum of the first n integers? If not, see Integrals post;)

Another idea, (which Integral provided while I was typing) write your sum as: $$\displaystyle\sum\limits_{i=1}^{n}(2i-1)$$ and fiddle with it to get an expression involving $$\displaystyle\sum\limits_{i=1}^{n}i$$, and apply the formula for this (again, using the ideas in Integral's post if need be). This would be a more easily applied general idea that can handle any arithmetic series $$\displaystyle\sum\limits_{i=0}^{n}(ki+d)$$.

Last edited: Sep 12, 2004
6. Sep 13, 2004

HallsofIvy

Staff Emeritus
Since you are adding only odd numbers, one way to do is this (Assuming you know that the sum of the first n integers is n(n+1)/2):
The sum of all numbers from 1 to 2n is (2n)(2n+1)/2= n(2n+1). Now subtract off the sum of all even numbers from 2 to 2n: 2+ 4+ 6+ ...+ 2n= 2(1+ 2+ 3+ ... + n)=
2(n)(n+1)/2= n(n+1).

The sum of all odd numbers from 1 to 2n-1 is n(2n+1)- n(n+1)= n(2n+1- n- 1)= n2 just as shmoe said.

In particular, the sum of all odd numbers from 1 to 99 is: (99= 100- 1= 2(50)- 1)
502= 2500.

7. Sep 13, 2004

Gokul43201

Staff Emeritus
Also, here's a general rule that can be used in times of trouble :

If a series S, has terms t(n), which are polynomials in n, of order p, the sum S(n) is a polynomial of order p+1. The coefficients of this polynomial can be determined simultaneously, from the given p+2 terms in the series.

Example : If the series S, has terms, t(n) which are linear in n, ie : t(n) = an + b, then the sum, S(n) is a quadratic in n, ie: S(n) = An^2 + Bn + C

NOTE : The order of the polynomial can be determined by the method of "successive differences". If the p'th successive differences are equal, the order is p.

Example : Consider the sequence : 2, 6, 12, 20, 30, ...
The first differences are : 4, 6, 8, 10, ...
The second differences are : 2, 2, 2, ...

Since the second differences are equal, t(n) is a quadratic in n, ie : t(n) = an^2 + bn + c.

So you can assume the sum has a cubic form : S(n) = An^3 + Bn^2 + Cn + D
Plugging in the for S(1), S(2), S(3) and S(4) gives you 4 simultaneous equations in A, B, C, and D, which you can solve.