Why is $\sum_{j=1}^{n}(n-j)$ equal to $\frac{n(n-1)}{2}$?

Question

As stated in the title, why is $\sum_{j=1}^{n} (n-j)$ equal to $\frac{n(n-1)}{2}$?

I know it's a very basic example of a sum but I'm currently a bit stumped and don't really know where to go from here..

Answer 1

$$\sum_{j=1}^n (n-j)=$$ $$\sum_{j=1}^nn-(1+2+3+...+n)=$$ $$n^2-\frac {n (n+1)}{2}=$$ $$\frac {2n^2-n^2-n}{2}=$$ $$\frac {n (n-1)}{2} .$$

Answer 2

Notice that as $j$ varies from $1$ to $n$, the value $n-j$ varies from $n-1$ to $0$: $$\begin{array}{c|c|c|c|c|c|c|c} j & 1 & 2 & 3 & \cdots & n-2 & n-1 & n \\ \hline n-j & n-1 & n-2 & n-3 & \dots & 2 & 1 & 0\end{array}$$

So letting $k=n-j$, and re-indexing the sum, we get $$\sum_{j=1}^n (n-j) = \sum_{k=0}^{n-1} k$$

Can you take it from here?

Answer 3

The claim is that

$(n-1)+(n-2)+\cdots+ 3+2+1+0=\frac{n(n-1)}{2}.$

Start pairing up the first and last elements:

\begin{align} (n-1)+1&=n\\ (n-2)+2&=n\\ (n-3)+3&=n\\ &\vdots \end{align}.

Now, if $n-1$ is even, then there are exactly $(n-1)/2$ of these "pairs," so we have a total of $(n-1)/2$ times $n$.

Otherwise $n-1$ is odd, so we can do a similar thing to count that $1+\cdots+n=n(n+1)/2$ ($n/2$ copies of $(n+1)$).

Answer 4

Well, $(n-1) + ...... + 2 + 1 =$

$1 + 2 + ..... + (n-1)$.

And if $m = n-1$ we know that $1 + 2 + ...... + m = \frac {m(m+1)}2 = \frac {(n-1)n}2$.

So the only real question is why does $1 + ..... + m = \frac {m(m+1)} 2$?

The easiest way to see that is:

$$1 + 2 + 3 + ....... + m = \sum\limits_{i=1}^m i\\ m + (m-1) + (m-2)+..... + 1 = \sum\limits_{i=1}^m i\\ -------\text{add those two together}-----------\\ (m+1) + (m+1) + ..... + (m+1) = \sum\limits_{i=1}^m i+\sum\limits_{i=1}^m i\\ m(m+1) = 2\sum\limits_{i=1}^m i\implies\\ \sum\limits_{i=1}^m i = \frac {m(m+1)}2$$

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There always more than one way to do things:

$\sum_{j=1}^n (n-j) = (\sum_{j=1}^n n)- (\sum_{j=1}^n j) = n*n - \frac {n(n+1)}2=n^2 - \frac {n^2}2 - \frac n2 = \frac {n^2}2 - \frac n2 = \frac {n^2 -n}2 = \frac {n(n-1)}2$