How can you easily add the numbers from 1 to 100?

[_Mayday_]

Hey, I've been trying to work this out for a few days now. There is supposed to be an easy way to add all the numbers from 1 to 100 easily, anyone know?

 

[Werg22]

I would venture to say that almost anyone on this forum knows how to sum an arithmetic progression. But here's a tip that doesn't call on much more than arithmetic. Consider adding the numbers from 1 to 100 twice. By cleverly arranging the addition like so

1 + 2 + 3 + 4 ... + 100 +
100 + 99 + 98 + 97 ... + 1

what constant relationship do you see between two numbers of the same column?

 

[Curvation]

I think that the easy way is:

1+2+3+...+100=x (1)
100+99+98+...+1=x (2)

combine this two (1)+(2)

101+101+101+...+101=2x
x=(100*101)/2=5050

 

[_Mayday_]

Brilliant cheers. I thought it had something to do with that but I struggled to put it into a formula.

 

[K.J.Healey]

I like to match up 100 with 0 myself, and you get all the way down to:
99+1, ... 51+49, then 50. So 100*50 + 50.

 

[country boy]

There is another way to visualize this using square arrays of elements. In a square array that is m=n+1 elements on a side, the number of off-diagonal elements in the lower left (or upper right) is (m^2-m)/2. But this is also the sum of the number of elements in all of the rows in that part of the matrix: 1+2+3+...+n. In our case n=100, m=101 and (m^2-m)/2=5050. This is equivalent to the Werg22 and Curvation solution.

To get K.J.Healey's solution, note that if we subtract half of the off-diagonal elements (say the upper right) in an nxn array then the remaining number of elements is n^2-(n^2-n)/2=(n^2+n)/2, again the sum of rows is 1+2+3+...+n. But this is n*(n/2)+(n/2) and is equal to 100*50+50 in our case.

Putting m=n+1 into (m^2-m)/2 gives (n^2+n)/2, so the two array visualizations are equivalent, of course.

 

[Alex48674]

I know how to do this but how would you do it if it weren't just intigers like 1 to n. What if you wanted to add 1+1/n where n = a positive intiger? let's say where n is 1 to 100

 

[a2tha3]

99+1 98+2 97+3 etc... 100+0 you take half of the number, and multiply it by 100, then add the number (5050)

1000*500+500= 500500 etc.

 

[Schrodinger's Dog]

Brilliant cheers. I thought it had something to do with that but I struggled to put it into a formula.

It's easy to see why this works if you represent each number as a total of individual numbers.

ie.

100000
110000
111000
111100
111110
111111

And so on, what do you notice about the ratio of 0's to 1's.

Incidentally there's a famous story where Gauss was asked this in a maths class at the age of 7, thinking to stump the children; Gauss gave him an immediate answer, something like the above answers given by the others, by figuring the above simple relation.

 

[ManDay]

You are all geniuses. Congratulations. I've you had lived about a 120 years ago you would have had developed the normal distribution, perhaps...

Edit: Aw Schroedinger, you were quicker than me. But it's no disgrace to be outran by a genius such as Schroedinger himself.

...or simply by his dog

(just kidding)

 

[kmarinas86]

Hey, I've been trying to work this out for a few days now. There is supposed to be an easy way to add all the numbers from 1 to 100 easily, anyone know?

Sure.

Think of a staircase. The first stair has a height of one. The second stair has a height of two and so on. The stair case forms a rough triangle, whose area is proportional to the square of the stair length. To make a long story short, the simple formula is:

1+2+3+...+number of stairs=(number of stairs^2+number of stairs)/2

number of stairs = 100

(100*100+100)/2

10100/2

5050

 

[Einstienear]

There is another way to visualize this using square arrays of elements. In a square array that is m=n+1 elements on a side, the number of off-diagonal elements in the lower left (or upper right) is (m^2-m)/2. But this is also the sum of the number of elements in all of the rows in that part of the matrix: 1+2+3+...+n. In our case n=100, m=101 and (m^2-m)/2=5050. This is equivalent to the Werg22 and Curvation solution.

To get K.J.Healey's solution, note that if we subtract half of the off-diagonal elements (say the upper right) in an nxn array then the remaining number of elements is n^2-(n^2-n)/2=(n^2+n)/2, again the sum of rows is 1+2+3+...+n. But this is n*(n/2)+(n/2) and is equal to 100*50+50 in our case.

Putting m=n+1 into (m^2-m)/2 gives (n^2+n)/2, so the two array visualizations are equivalent, of course.

Ime sure the poster to the thread said "easy".

Not that its hard or anything, but considering the amount of words the poster put in this post, the poster wants an easy solution

(relativity)

Einstienear

 

[NightGale]

Consider using arithmetic progression's formula:

n/2(a+n) where a is the 1st term = 1, and n is the number of terms = 100

 

[Air]

1+100=101, 2+99=101, 3+98=101 etc. There are 50 pairs of these so it is 101 times 50 which equals 5050.

 

[Gokul43201]

I know how to do this but how would you do it if it weren't just intigers like 1 to n. What if you wanted to add 1+1/n where n = a positive intiger? let's say where n is 1 to 100

\sum_{n=1}^{100}1+\frac{1}{n}=\sum_{n=1}^{100}1+\sum_{n=1}^{100}\frac{1}{n}
~~~~~=100+ \sum_{n=1}^{100}\frac{1}{n}
The remaining term above is a partial sum of a harmonic series. There is no simple, closed-form expression for such a partial sum to arbitrary number of terms. I think there are "compact expressions" for small partial sums and approximations for large ones, but that's all.

 

[Werg22]

An interesting consequence of being able to write down a formula for the sum of the first n naturals: we can write down a formula for the sum of the first n powers m, where m is any natural. To do so, we simply write

\sum_{k=0}^{n-1} (k+1)^{m+1}=\sum_{k=0}^{n} k^{m+1}

then\sum_{k=0}^{n-1} (k+1)^{m+1} - k^{m+1} = n^{m+1}

Of course, on the left, after a binomial expansion of the first term, we get rid of k^(m+1), left with with the sum of terms with the highest degree being m. So, if we know the formulas for the sums for the powers 1 to m-1, we determine the formula for m.

Of course, it's not very good-looking as you go up... but it's still interesting.

 

[Werg22]

\sum_{n=1}^{100}1+\frac{1}{n}=\sum_{n=1}^{100}1+\sum_{n=1}^{100}\frac{1}{n}
~~~~~=100+ \sum_{n=1}^{100}\frac{1}{n}
The remaining term above is a partial sum of a harmonic series. There is no simple, closed-form expression for such a partial sum to arbitrary number of terms. I think there are "compact expressions" for small partial sums and approximations for large ones, but that's all.

There's also the approximation log(100) + {\gamma}

With {\gamma} = 0.577215665...

I wonder if it's possible to measure the order of magnitude of the error.

 

[BryanP]

Just think about the series/sequence. I remember learning about this in an introductory calculus class when we started learning about sequences and series. From there, you can generate formulas such as these.

 

[J R]

Hey, I've been trying to work this out for a few days now. There is supposed to be an easy way to add all the numbers from 1 to 100 easily, anyone know?

100 X 101 divided by 2 = answer. All numbers work the same. 50 x 51 divided by 2
44 x 45 divided by 2 10 x 11 divided by 2

 

[J R]

numbers 1 to100

100 X 101 divided by 2 = answer. All numbers work the same. 50 x 51 divided by 2
44 x 45 divided by 2 10 x 11 divided by 2

In other words the number you are adding times the next highest number divided by 2 = answer

 

[rostbrot]

same thing as what Air said, but it just looks so pretty written out!

1______+_______100_= 101
_2_____+______99___= 101
__3____+_____98____= 101
__...___...____...____...
___49__+__52_______= 101
____50_+_51________= 101

so you got 101 50 times 101x50=5050

so... when summing numbers 1 to N
sum = (N+1)(N)/2

which simplifies out to the same thing JR said.

 

[d1ff30m0rf1zm]

Hi. A simple solution:

Let's look at the case n = 5 and write it out like this:

O
O O
O O O
O O O O
O O O O O

Immediately we see that we can fill it in like this:

O = = = =
O O = = =
O O O = =
O O O O =
O O O O O

So we've got a 5 x 5 square. Ignore the last row for now:

O = = = =
O O = = =
O O O = =
O O O O =

This "thing" is symmetric along the diagonal. Hence, it seems like we could simply find the total number of elements and divide by 2. That is the case: (1/2)(4 x 5) = 10. Counting, you'll see that its correct. Note that we've added an auxiliary column to make this work. This suggests that for n = 5 we simply add a sixth column to make the array symmetric:

O = = = = =
O O = = = =
O O O = = =
O O O O = =
O O O O O =

Evidently this is 5 x 6, or more generally n(n + 1). Cutting in half gives us the elements we are looking for.

So it seems like starting with n = 1 can't give us a symmetric array without adding an auxiliary column. So let's begin with n = 0. The "." is a placeholder.

.
O
O O
O O O
O O O O

Now this allows us to fill in with 4, 3, 2, and 1 extra elements on each row:

= = = =
O = = =
O O = =
O O O =
O O O O

We have 5 rows now. The formula follows.

 

[d1ff30m0rf1zm]

I've been messing around with the above method and it *seems* like the formula for the sum of the first n integers in the form \displaystyle\frac {m(m + 1)}{2} is

\displaystyle \frac {n^2(n + 1)}{2} - \left[\sum_{k = 0}^{n - 1} (n - k)(n - k - 1) \right].

Not sure. I think its more or less completely useless. I can show my work if anyone is interested.

 

[phoenixankit]

I know this topic is more than a month old, but I just couldn't help answering

The formula is:
S= n/2 [2a + (n-1)d]

Where,
S is Sum of numbers
n is number of numbers
a is the starting number
d is the difference between the first two numbers

in this case,
S=100/2[2x1 + (100-1)1]
=50[2+99]
=50 x 101
=5050
So, S=5050

 

[robphy]

Incidentally there's a famous story where Gauss was asked this in a maths class at the age of 7, thinking to stump the children; Gauss gave him an immediate answer, something like the above answers given by the others, by figuring the above simple relation.

Here are some interesting articles on this "famous story"...

"Gauss's Day of Reckoning"
Brian Hayes
May-June 2006, Volume 94, Number 3, Page: 200, DOI: 10.1511/2006.59.3483
http://www.americanscientist.org/issues/pub/gausss-day-of-reckoning

"Versions of the Gauss Schoolroom Anecdote"
Brian Hayes
http://www.sigmaxi.org/amscionline/gauss-snippets.html

 

[DeaconJohn]

I know how to do this but how would you do it if it weren't just intigers like 1 to n. What if you wanted to add 1+1/n where n = a positive intiger? let's say where n is 1 to 100

Good question!

They're called the harmonic numbers, or the finite harmonic series. I don't know a formula for them, though I would like to.

For large n, you can get an approximate answer by taking the logarithm of n to the base e and then adding the Euler-Mascheroni constant. The error in this approximation goes to zero as n goes to infinity.

Reference: Wikipedia.

http://en.wikipedia.org/wiki/Euler-Mascheroni_constant

DJ

P.S. This is assuming that you meant to write "add 1 to 1/n where n = a positive intiger"
as opposed to "add 1+1/n where n = a positive intiger" which is what you actually wrote.

 

[DeaconJohn]

There's also the approximation log(100) + {\gamma}

With {\gamma} = 0.577215665...

I wonder if it's possible to measure the order of magnitude of the error.

Werg22 - I wonder too. DJ

 

[DeaconJohn]

There's also the approximation log(100) + {\gamma}

With {\gamma} = 0.577215665...

I wonder if it's possible to measure the order of magnitude of the error.

The Wikipedia section on Asymptotic Expansions in the Harmoinc Numbers article implies the expression (1/2n) as an estimate of the error.

Mote accurately, it gives (1/2n) as the first term in the "Euler" expansion for the error.

It gives two other asymptotic expansions for the error.

It says the third, the Caesaro or Ramanujan expansion, gives a [supposedly more accurate] expansion for the error between the average of two successive parital sums of the harmonic series and the approximation you quote.

http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

 

[DeaconJohn]

Consider using arithmetic progression's formula:

n/2(a+n) where a is the 1st term = 1, and n is the number of terms = 100

NightGale,

Or, maybe, n(a+b)/2 where a (= 1) is the 1st term, b (= n) is the last term, and n is the number of terms?

Derivation of the formula n(a+b)/2:

The sum from "a" to "b", a+...+b, has b-a+1 terms. So, n = b-a+1.

Take the sum from one to "b" and subtract the sum from one to "a-1." You are left with the sum from "a" to "b." So,

b(b+1)/2 - (a-1)a/2 = (b^2 - b - a^2 + a)/s = ... = (b-a+1)(a+b)/2 = n(a+b)/2.

DJ

P.S.

I got the correction to your formula from Wikipedia. However, I did not like their derivation so I came up with my own.

It was not obvious to me that the expressions on either side of the dots ("...") were equal to each other. It was a pleasant surprise to see how the algebra worked out.

I omitted the equalities represented by the dots ("...") in the above sequence of equalities so as not to rob you of your own pleasant surprise.

 

[Moses188]

I was asked this question at trivia night at my uni bar, and figured out a rather layman's way of doing it quickly:
1+2+...+8+9=45
So 10 through 19 will just be 10x10+45=145, the 20s are 20x10+45=245 and so on, adding one hundred to each successive set of ten till 99, then add 100 at the end = 5050.

There were no prizes for third that night but atleast I learned something :)