1+2+3+...+n=(1+n)+(2+(n-1))+(3+(n-2))+...
=(1+n)+(1+n)+(1+n)+...
How many copies of n+1 are in there in the second line?
with n=6 and n=7
Since the n numbers are divided up in pairs, I'd say n/2 copies.
Hence the formula for the sum: n(n+1)/2
Technically, that's [n+1]/2 where [] is the greatest-integer function.
To determine the number of copies of (n+1) in the second line, we need to express the second line in a clearer form.
Let's evaluate the second line for n=6 and n=7:
For n=6:
(1+6)+(2+(6-1))+(3+(6-2))+(4+(6-3))+(5+(6-4))+(6+(6-5))
=7+7+7+7+7+7
=42
For n=7:
(1+7)+(2+(7-1))+(3+(7-2))+(4+(7-3))+(5+(7-4))+(6+(7-5))+(7+(7-6))
=8+8+8+8+8+8+8
=56
In the second line, we can see that for both n=6 and n=7, there are n copies of (n+1) present. So, the number of copies of (n+1) in the second line is equal to n.
Therefore, for n=6, there are 6 copies of (n+1) in the second line.
For n=7, there are 7 copies of (n+1) in the second line.