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What is the term for a factorial type operation, but with summation instead of products?

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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    @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.2011-11-04

2 Answers 2

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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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    @ArturoMagidi$n$ Edit: that should go that you've paired the first term with the last, the second with penultimate, and so on, except, if it exists, the exact middle term which you've paired with itself once, while I haven't paired it with anything, or duplicated pairs.2011-11-04
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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.