# Finding a formula to sum natural numbers up to $n$ [duplicate]

I got this question in homework:

Find an expression for the sum ‫‪

$\sum k = 1 +\cdots + n‬‬$.

and prove it using an induction.

I'm not even near finding the expression. What I did notice is that if $n$ is (for example) 5 then the sum would be

$5^2 - 4^2 + 3^2 - 2^2 + 1^2$

So the first number is always positive and from there on the sign changes.

Any tips on how do I contintue from this point on, assuming I'm in the right direction?

Thanks!

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## marked as duplicate by J. M., Henning Makholm, Srivatsan, Asaf Karagila, mixedmath♦Nov 4 '11 at 15:16

maybe this could be helpful – pedja Nov 4 '11 at 14:09
@yotamoo: This is a very nice observation that can be used to find the summation formula for the squares starting from the summation formula you are trying to prove at the moment. – Phira Nov 4 '11 at 14:32

Hint. $$\begin{array}{cccccccccccc} & 1 & + & 2 & + & 3 & + & 4 & + & \cdots & + & n\\ +& n & + &n-1 & + & n-2 & + & n-3 & + & \cdots & + & 1\\ \hline & n+1 & + &n+1 & + &n+1 &+& n+1 & + & \cdots & + & n+1 \end{array}$$

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To expand on this, if n is even, how many instances of (n+1) do we have here? – Doug Spoonwood Nov 4 '11 at 14:06
@Doug You do not need $n$ to be even for this answer. There are always $n$ occurrences of $(n+1)$ in the last line, odd or even. – Srivatsan Nov 4 '11 at 14:12
@all: This is often called Gauss' trick. – Srivatsan Nov 4 '11 at 14:13
I love this trick -- usually blows the mind of at least one student when shown to them. Can you do something similar for the sum of squares, cubes, etc? – Mike Wierzbicki Nov 4 '11 at 14:14
@Srivatsan Huh? If n=4, then we have 1+2+3+4, so we have 4+1, and 4+1 in the last line. So, we have 2 occurrences of (n+1) in the last line. If n=5, then we have 1+2+3+4+5, so we have 4+1, 4+1, and 3 for 2 occurrences of (n+1) in the last line. You're right n doesn't need to be even for this answer, but the question I asked seems simpler to answer if you only consider the even cases. – Doug Spoonwood Nov 4 '11 at 14:25

Hint:

If $S_n$ is your sum $S_n=1+2+3+\ldots+n$

then, it's simple you see that

$S_{n}-S_{n-1}=(1+2+\ldots+n-1+n)-(1+2+\ldots+n-1)=$

$(1-1)+(2-2)+\ldots+(n-1 -(n-1))+n=n$.

Now you can use a trick, you can think $S_n$ is a polynomial of degree 2, so

$S_n=an^2+bn+c$

Putting this in the equation

$S_n-S_{n-1}=n$, you will find $a$, $b$ and $c$ and then, you get your sum or $S_n=an^2+bn+c=1+2+3+\ldots+n$.

This is more general, but @ArturoMagidin ways is the easiest.

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