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Can anyone explain to me how the following two summations are equal?

$=\sum_{j=i+1}^n j$

$=\sum_{j=1}^n j - \sum_{j=1}^i j$

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  • $\begingroup$ Just write out the sums explicitly: $\sum_{j=1}^n j = 1 + 2 + 3 + \dots + i + (i+1) + \dots + n$ etc. and it should just pop. BTW, there's a hidden assumption in your question that $i \le n$. You should check how violating this affects the sums. $\endgroup$
    – NickD
    May 30 '17 at 2:27
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a rule that helps identify how to expand it from the first summation to the second summation

The basic rule that applies in these cases is that, if $a \le b \le c$ then:

$$ \sum_{j=a}^c f(j) = \sum_{j=a}^b f(j) + \sum_{j=b+1}^c f(j) $$

The rule simply says that when you add the $c-a+1$ terms $f(a), f(a+1), \cdots,f(c)\,$, you get the same result if you split the sum in two, which follows directly from the associativity of addition:

$$f(a)+ f(a+1)+ \cdots+f(c) = \big(f(a) + f(a+1) + \cdots + f(b)\big) + \big(f(b+1) + \cdots + f(c)\big)$$

The rule can obviously be rewritten as: $$ \sum_{j=b+1}^c f(j) = \sum_{j=a}^c f(j) - \sum_{j=a}^b f(j) $$

The latter reduces to the posted question for $\,a=1, b=i, c=n\,$ and $\,f(j)=j\,$.

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$$ \color{#c00}{(i+1)+(i+2)+\cdots+n} = [\color{#0a0}{1+2+\cdots+i}+\color{#c00}{(i+1)+(i+2)+\cdots+n}]-[\color{#0a0}{1+2+\cdots+i}] $$

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  • $\begingroup$ Tnx for your help. That makes sense. Can you tell if there is a rule that helps identify how to expand it from the first summation to the second summation? $\endgroup$
    – jason
    May 30 '17 at 2:33
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$$S_1=\sum_{j=i+1}^n j=(i+1)+(i+2)+\cdots+n$$ $$\require{cancel}S_2=\sum_{j=1}^n j - \sum_{j=1}^i j=\left[\cancel{1+\cdots+{(i-1)}+ i}+(i+1)+\cdots+n\right]-[\cancel{1+\cdots+{(i-1)}+ i}]=S_1$$

The first sum $S_1$ is saying to add all the numbers starting from $i+1$ to $n$, the second sum $S_2$ is saying to add all the numbers from $1$ to $n$, but then take away all the numbers from $1$ to $i$. This would then leave behind just the numbers from $i+1$ to $n$, as the first sum said.

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  • $\begingroup$ Tnx for your help. That makes sense. Can you tell if there is a rule that helps identify how to expand it from the first summation to the second summation? $\endgroup$
    – jason
    May 30 '17 at 2:33
  • $\begingroup$ @jason I am not sure if I understand you correctly - for this sum, it is just the sum of $j$ from $i+1$ to $n$. Notice how I read that off from the notation. Having interpreted it like this, the meaning becomes clear - you just add each number up from $i+1$ to $n$. $\endgroup$
    – John Doe
    May 30 '17 at 2:43

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