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Can someone show a step-by-step process for simplifying the summation to $i \cdot n - i + 1$ as shown:

$$\sum _{j=i}^{i \cdot n} 1 = i\cdot n - i + 1$$

I don't know how to begin to solve this.

Wolfram Alpha

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up vote 3 down vote accepted

We are adding up $in-i+1$ terms, all equal to $1$. What happens when we add $47$ $1$'s together?

It is easy to make an error and be off by $1$. For example, what is $\sum_{j=2}^6 1$? We have a term of $1$ for each of $j=2,3,4,5,6$. The number of terms is not $6-2$, it is $6-2+1$.

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I suppose I'm having trouble reading the summation notation. Please correct me if I am wrong: In my case, i is some constant and j is set to i. i*n is the upper limit for j. If i is set to 47, then j = 47 and the limit is 47 * n so it would run 47 * n - i + 1 times.... MUAHAHAAH... thanks your awesome! –  James T Apr 30 '12 at 5:34
    
@JamesT: Paradoxically, a sum like $\sum_{j=i}^{in} j$ is easier to understand than $\sum_{j=i}^{in} 1$ or $\sum_{j=i}^{in} 7$, even though the first one takes some work to evaluate, while the last two are very easy. Your summary was right. –  André Nicolas Apr 30 '12 at 5:47
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