Why is $\sum_{j=i+1}^{n-1}1 = -i+n-1$?

Question

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 $1$?

Answer 1

$$\sum_{j=1}^{n-1}1=n-1$$ and $$\sum_{j=1}^{i}1=i$$ notice that $$\sum_{j=i+1}^{n-1}1=\sum_{j=1}^{n-1}1-\sum_{j=1}^{i}1=n-1-i$$

Answer 2

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

$$\sum_{j=1}^{3}1 = 1+1+1 = 3-1+1=3$$

Answer 3

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.

Answer 4

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. $n-i-1$.

Answer 5

$j=i+1 , j=i+2 , j=i+3,. ..., j= n-1 = i + (n-1 -i).$

You count from $1$ up to $(n-1-i)$, which are $(n-1-i)$ terms.

The sum is ?

Answer 6

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.