Number of numbers between a and b and sums from x to y This is for my benefit and curiosity and not homework.
How do you calculate the number of numbers between $1$ and $100$? How do you calculate the number of even and odd numbers between $1$ and $100$?
How can you use the grouping method (e.g. sum from $1$ to $100$ = $(1+101)x50$ or $(2+99)50))$ to find the sum of numbers from $5$ to $n$? If you're not familiar with the grouping method can you explain the intuition behind the formula for sums from $1$ to $n$: $(n(n+1)/2)$?
What is the sum of numbers from $5$ to $n$?  
 A: 
What is the sum of numbers from 5 to n? 

$$\sum_{k = 1}^n k - \sum_{k=1}^4 k = \frac{n(n+1)}2 - \frac{4(4+1)}{2} = \frac{n(n+1)}{2} -10$$ 
A: 
Explain the intuition behind the formula for sums from 1 to n:
  (n(n+1)/2).
What is the sum of numbers from 5 to n?

Between 5 and n (inclusive) there are C = (n - 5 + 1) numbers.
The average of those numbers is: M = (5 + n) / 2
We can safely multiple the number of numbers by the average, since the distribution is linear:
sum = C * M = (n - 5 + 1) * (5 + n) / 2
If it does not seem immediately intuitive then try a geometric proof:


*

*draw a graph of the numbers between 5 and n, against their values (y=x for x in {5..n}).

*Agree that the sum you are looking for is the area under the graph.

*Draw a horizontal line through the mean value (5+n)/2.

*Now looking at the graph, in the n column, remove all the boxes above the average, and put them on the left end of the graph (at 5).  This should bring the height of column 5 up to the average line.  Now do the same for (n-1) and 6, then for (n-2) and 7.  Eventually you can get every column to fit the average height.

*It should now be clear that the area under the graph is the mean value multiplied by the number of columns.
Fortunately I am a budding artist, so I can present this explanation visually:

The blue line is the average, or mean value.  We start with a wedge and end with a rectangle.
I suppose one could rotate the whole triangle above the mean line in one go.  But the way I have always considered it laboriously moving boxes from column n into column 5, then from n-1 into 6, and so on...
A: Alice and Bob started work on the same day. Alice's wage the first day was $5$ dollars, the next day (she is a good worker) it was $6$ dollars, the next day it was $7$ dollars, and so on. On the last day she worked, she earned $n$ dollars. 
Note that this means her total income $A$ was given by 
$$A=5+6+7+\cdots+n\tag{1},$$
and she worked for $n-5+1$ days.
Bob's wage the first day was $n$ dollars. But the next day his wage was decreased by a dollar, and the same happened the day after that, and so on. On his last day he got $5$ dollars. 
It is clear that Bob's total income was also $A$.
I forgot to mention that Alice and Bob are "partners." Every day, their joint income was $n+5$, since that was their joint income the first day, and every day Alice's income went up by $1$, and Bob's went down by $1$, leaving their combined daily income unchanged.
Between them, they earned $2A$ dollars. Every one of the $n-4$ days, they earned a combined $n+5$ dollars, so
$$2A=(n-4)(n+5).$$
It follows that
$$A=\frac{(n-4)(n+5)}{2}.$$
This gives us the desired closed form expression for the sum (1).  
Remark: The same "story" can be used to find a closed form for the sum $a+(a+d)+(a+2d)+\cdots +(a+md)$. 
A: If you want to calculate the sum: $$a_1+a_2+ \dots+ a_N$$
you can use the formula:
$$\frac{N \cdot (a_1+a_N)}{2}$$
where $N$ is the number of terms of the sum.
In your case,you take $a_1=5$, $a_N=n$ and $N=n-5+1$
So,the sum is equal to :
$$\frac{(n+5)(n-4)}{2}=\frac{n(n+1)}{2}-10$$
