I'm having trouble understanding summation, specifically the following example:
$\sum^{n}_{j=n-i+1} 1$
If I plug this into WolframAlpha, I find that it sums to $i*n$, but I don't understand how to work through the problem myself. Since it's summing 1 within the specified indexes, I tried writing it out as $(n-i+1) + (n-i+1) + ... +(n-i+1)= n(n-i+1)$ but I don't think this is the right way to approach it. Any help would be greatly appreciated.
The sum you have written, $$\sum_{j=n-i+1}^{n}1$$ is interpreted as $$\underbrace{1+1+...+1}_{n-(n-i+1)+1 \mbox{ times}} = \underbrace{1+1+...+1}_{(i-1)+1 \mbox{ times}}=\underbrace{1+1+...+1}_{i \mbox{ times}}=i$$
To give you a better example, we could write $$ 4 = 1+1+1+1 = \underbrace{1+1+1+1}_{(4-1)+1 \mbox{ times}}=\sum_{j=1}^{4} 1$$ This is a special case of your question, with $n=4$ and $i=4$