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I'm having trouble with this equation mainly because it has a couple of odd things with it, and its these that have thrown me off as i'm not to sure how to tackle them. The equation is:

n - 1
SUM    ( n - i) = n(n - 1)/2
i = 1

Heres my method of working it out so far:

Assume a baseline of i = 1.
(1 - 1) on LHS = 0
1(1 - 1)/ 2 on RHS = 0
Thus the equation can be solved.

Here we test if it also works with (n + 1) which would be (n - 1 + 1)? = n.
LHS:
So (n - i) = ((n + 1) + n) = 2n + n
n(n - 1)/2 + (2n + n)
    = (n^2 - n + 6n) / 2

RHS:
Basically add 1 to every n available.
(n + 1)((n + 1) - 1)/2
= (n^2 + n)/2

Thus, Since LHS != RHS Its not true?

I'm not really to sure. Here are my main problems:

With the (n - 1) at the top of the SUM, do I add 1 for every iteration or minus 1 so it becomes (n - 2)?

Also, with the (n - i), for the LHS what does that become? because I dont think (2n + n) is correct.

Thanks for all the help! I've been stuck on this one for a while with all the others being pretty simple kinda. :)

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I'm assuming you want to show that $$ \sum_{i=1}^{n-1}(n-i)=\frac{n(n-1)}{2}\tag{1} $$ holds for $n\in \{2,3,\ldots\}$ by using induction.

The induction start is to check that $(1)$ holds for $n=2$. But the LHS is $$ \sum_{i=1}^1 (2-i)=2-1=1 $$ and the RHS is $\frac{2(2-1)}{2}=1$ so we see that $(1)$ is satisfied for $n=2$.

Now, the induction step is to assume that $(1)$ holds for some $n\in \{2,3,\ldots,\}$ and then we aim to show that $(1)$ also holds for $n+1$. The assumption that $(1)$ holds for $n$ means exactly that $$ \sum_{i=1}^{n-1}(n-i)=\frac{n(n-1)}{2}, $$ and we want to show that $$ \sum_{i=1}^{(n+1)-1}((n+1)-i)=\frac{(n+1)((n+1)-1)}{2}\tag{2} $$ Here I have just plugged in $n+1$ instead of $n$ in the formula for $(1)$. But rewriting $(2)$ we have to show $$ \sum_{i=1}^n(n+1-i)=\frac{n(n+1)}{2}\tag{3} $$ So try and start out with the left-hand side of $(3)$ and rewrite it so that you can use your assumption (that is $(1)$). I hope this helps.

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