Adding consecutive numbers

Frank Spencer

What is the simplest way to add 1 plus 2 plus 3 plus 4 etc. all the way to 2000?

Is there a formula that can be used or can it be done in excel?

rainbow kirby

n(n+1) / 2 should give you the answer

Fremen

1 + 2 + 3 + ... + 1999 + 2000 = x
2000 + 1999 + ... + 3 + 2 + 1 = x

Now add corresponding numbers in each sequence and you get 2000 copies of the number 2001, which you know is equal to 2x.

Seems a bit homework-ish to me.

hivizman

There's a famous story about the mathematician Carl Friedrich Gauss (1777-1855) that, when he was in primary school, his teacher wanted to keep the class busy for a period and told them to add together all the numbers from 1 to 100. Although the teacher expected this to take the boys a considerable amount of time, Gauss produced the answer within a couple of minutes, using the method that Fremen outlined. Although Wikipedia suggests that this story may be apocryphical, it's certainly consistent with Gauss's later career as a mathematician.

Eliot Rosewater

I was highly bored, so I decided to do a formal proof of this :pac:

Proposition:

[latex]\displaystyle \sum_{i=1}^n i=\frac{n(n+1)}{2}[/latex]

Proof by induction.
Case i=1

[latex]\displaystyle \sum_{i=1}^1 i=\frac{1(1+1)}{2}[/latex]

[latex]\displaystyle \sum_{i=1}^1 i=\frac{2}{2}[/latex]

[latex]\displaystyle 1=1[/latex]


Assume true for n=k, so that

[latex]\displaystyle \sum_{i=1}^k i=\frac{k(k+1)}{2}[/latex]


Now show that its true for n=k+1

[latex]\displaystyle \sum_{i=1}^{k+1} i=\left(\sum_{i=1}^{k} i\right) + \left(k+1\right)[/latex]

[latex]\displaystyle \sum_{i=1}^{k+1} i=\left(\frac{k(k+1)}{2}\right) + \left(k+1\right)[/latex]

[latex]\displaystyle \sum_{i=1}^{k+1} i=\left(\frac{k(k+1)}{2}\right) + \left(\frac{2(k+1)}{2}\right)[/latex]

Taking out a common factor, (k+1)

[latex]\displaystyle \sum_{i=1}^{k+1} i= \frac{(k+1)(k+2)}{2}[/latex]

[latex]\displaystyle \sum_{i=1}^{k+1} i= \frac{(k+1)((k+1)+1)}{2}[/latex]

True, etc...

LeixlipRed

A much simpler and more aesthetic proof I believe is this one:

[latex]S_{n} = 1+2+...+n-1+n[/latex]

and also

[latex]S_{n} = n+n-1+...+2+1[/latex]

Add the two lines to get:

[latex]2S_{n} = n(n+1)[/latex]
[latex]S_{n}=\frac{ n(n+1)}{2}[/latex]

f1dan

Frank Spencer wrote: »

What is the simplest way to add 1 plus 2 plus 3 plus 4 etc. all the way to 2000?

Is there a formula that can be used or can it be done in excel?

To do it quickly in excel, do this:

Type 1 into cell A1
Type 2 into cell A2
Highlight cells A1 and A2
Hover over the bottom right corner of cell A2 until the black cross appears
Click, hold and drag down until you get to A2000
In cell A2001, type in =SUM(A1:A2000) and you should get your answer

Eliot Rosewater

LeixlipRed wrote: »

A much simpler and more aesthetic proof I believe is this one

Its simpler for sure, but I myself don't like using "..." in proofs. I like my proofs to be mechanical leave no room for lack of intuition :pac:

So a better proof than my induction one, and using your methodology, would be:

[latex]\displaystyle 2S_{n}=\sum_{i=1}^n i+\sum_{i=1}^n i[/latex]

Reversing the direction of the second sequence we get

[latex]\displaystyle \sum_{i=1}^n i=\sum_{i=1}^n (n+1-i)[/latex]

So that

[latex]\displaystyle 2S_{n}=\sum_{i=1}^n i+\sum_{i=1}^n (n+1-i)[/latex]

[latex]\displaystyle 2S_{n}=\sum_{i=1}^n (n+1)[/latex]

Something summed n times, is that something times n. Dividing by two...

[latex]\displaystyle S_{n}=\frac{n(n+1)}{2}[/latex]

Fremen

Eliot Rosewater wrote: »

Its simpler for sure, but I myself don't like using "..." in proofs. I like my proofs to be mechanical leave no room for lack of intuition :pac:

Best get used to it. The further on you go, the sketchier they become

Mellor

f1dan wrote: »

To do it quickly in excel, do this:
....

To do it quicker, do it in your head.

Yakuza

hivizman wrote: »

There's a famous story about the mathematician Carl Friedrich Gauss (1777-1855) that, when he was in primary school, his teacher wanted to keep the class busy for a period and told them to add together all the numbers from 1 to 100. Although the teacher expected this to take the boys a considerable amount of time, Gauss produced the answer within a couple of minutes, using the method that Fremen outlined. Although Wikipedia suggests that this story may be apocryphical, it's certainly consistent with Gauss's later career as a mathematician.

He is also reputed to have said, (when his teacher wrote the numbers 1-10 on the board and asked "which of these numbers can be divided by 2"), "all of them"

There are several handy formulas for summing the varous series (n, n², n³ etc) (your case was a simple case of an arithmetic series with the start value equal to 1 and the increment equal to 1). If you google "arithmetic series" you'll get more info.