Why is $\sum_{j=i+1}^{n-1}1 = -i+n-1$?
I am having trouble understanding this result, we are summing up $1$ from $i+1$ to $n-1$ wouldn't that just be $1$?
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2Just check: How many summands are there? – Hagen von Eitzen Feb 20 '18 at 10:57
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$j=i+1 , j=i+2 , j=i+3,. ..., j= n-1 = i + (n-1 -i).$
You count from $1$ up to $(n-1-i)$, which are $(n-1-i)$ terms.
The sum is ?
Peter Szilas
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Note that by definition
$$\sum_{j=a}^{b}1 = b-a+1$$
that is the number of summands, then
$$\sum_{j=i+1}^{n-1}1 = n-1-(i+1)+1=n-1-i$$
To convince yourself let try with some simple example
$$\sum_{j=1}^{3}1 = 1+1+1 = 3-1+1=3$$
user
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Since by definition $$\sum_{i=1}^n f(i)=f(1)+f(2)+...+f(n)$$ for $f(i)=1$ we have $$\sum_{i=1}^n 1=1+1+...+1=n-1+1=n$$ and $$\sum_{i=a}^b 1=1+1+...+1=b-a+1$$ – user Feb 21 '18 at 01:27
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It's just the following: $$n-1-(i+1)+1=n-i-1$$
You can use $a_n=a_1+(n-1)d$ for the arithmetic progression: $$n-1=i+1+(x-1)\cdot1,$$ where $x$ is the needed sum.
Michael Rozenberg
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How can we explain this in words? We are starting at i+1 and ending at n-1 and how does the 1 come into play? – bigfocalchord Feb 20 '18 at 11:00
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You can use $a_n=a_1+(n-1)d$ for the arithmetic progression: $n-1=i+1+(x-1)1$ – Michael Rozenberg Feb 20 '18 at 11:02
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$$\sum_{j=1}^{n-1}1=n-1$$ and $$\sum_{j=1}^{i}1=i$$ notice that $$\sum_{j=i+1}^{n-1}1=\sum_{j=1}^{n-1}1-\sum_{j=1}^{i}1=n-1-i$$
cansomeonehelpmeout
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Replace $1$ with $a_j$, so your sequence is $a_{j+1}+a_{j+2}+\dots+a_{n-1}$.
Now let $a_j=1\;\;\forall j\in\mathbb{Z^+}$.
The sum becomes $1+1+\dots+1$, i.e. $n-i-1$.
JMP
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Note that summation goes from $(i+1)$ to $(n-1)$: $$\begin{align}\sum_{j=i+1}^{n-1} 1=&1_{i+1}+1_{i+2}+\cdots+1_{n-1}=\\&1_{i+1}+1_{i+2}+\cdots+1_{i+n-i-1}.\end{align}$$ Thus, there are $n-i-1$ numbers $1$ added.
farruhota
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