I understand that summing numbers between 1 to n with a difference of d=1 is expressed as $\sum_{k=1}^{n}k = 1 + 2+ ... + n = \frac{1}{2}n(n+1)$
I also understand that the above equation is a special case of the general arithmetic series equation of $\sum_{k=1}^{n}a_k = a_1 + a_2+ ... + a_n = \frac{1}{2}n(a_1+a_n)$
where $ a_1 = 1$ and $a_n = n$
So if I want to find the summing of numbers between 2 to n with difference d=1, is it possible to say $\sum_{k=2}^{n}a_k = a_2+ ... + a_n = \frac{1}{2}n(a_2+a_n)$
because $\sum_{k=2}^{n}a_k = a_2+ (a_2 + d) + (a_2 + 2d) + ... + (a_2+(n-1)d) $ and $\sum_{k=2}^{n}a_k = a_n+ (a_n - d) + (a_n - 2d) + ... + (a_n-(n-1)d) $
so adding the two equation, we get $2\sum_{k=2}^{n}a_k = n(a_2 + a_n) $
However, when I apply this equation I derived $\sum_{k=2}^{n}a_k = a_2+ ... + a_n = \frac{1}{2}n(a_2+a_n)$,
I get $\frac{1}{2}n(2+n)$ from substituting $a_n = n $ and $a_2 = 2$
as opposed to $\frac{n(n-1)}{2}-1$ which I know for a fact is correct because $\sum_{k=2}^{n}k = (\sum_{k=1}^{n}k) -1 = (\frac{1}{2}n(n+1)) - 1 \neq \frac{1}{2}n(2+n)$