Where does the -2 +1 come from in this sigma notation?

Question

I am solving this equation:

My issue arises on line 2 where we have (n + 1 - 2 + 1)(n + 1 + 2)/2. This is what i understand. We have the formula n! = n(n+1)/2. Subbing values into the equation yields us with 10[(n+1)(n+1 + 1)/2] however we need to account for the fact we are starting at j = 2. I would then do this by 10[(n+1-2)(n+1+1)/2]. What I dont understand is why in the solution they have given

• (n+1 - 2 - 1 +1) the -1 + 1
• (n + 1 + 2) the + 2

How did they make that logical jump and what were the steps/thought process in doing so.

Answer 1

in line 2 the $\frac{(n+1+2)}{2}$ part is the average of the terms being summed. In this problem this is calculated by adding the first and last term and dividing by 2. The $(n+1-2+1)$ part is the number of terms being summed. This is calculated by subtracting the index of the first term (2) from the index of the last term (n+1) and adding one.

The sum is equal to the average value of each term multiplied by the number of terms.

Also, $n!=\frac{n(n+1)}{2}$ is not true. The correct equation is $\sum_{i=1}^n i = \frac{n(n+1)}{2}$

Answer 2

$n+1-2+1$ gives the number of terms, which is the number of integers in the inclusive range $[2,n+1]$.

$\dfrac{n+1+2}{2}$ gives the average term, which is the largest term ($n+1$) plus the smallest term ($2$) divided by $2$.