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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 $\sum_{k=2}^{n}k = (\sum_{k=1}^{n}k) -1 = (\frac{1}{2}n(n+1)) - 1 \neq \frac{1}{2}n(2+n)$

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    $\begingroup$ \begin{eqnarray*} \sum_{k=2}^{n}a_k = a_2+ \cdots + a_n = \frac{1}{2} \color{red}{(n-1)}(a_2+a_n). \end{eqnarray*} $\endgroup$ – Donald Splutterwit Oct 13 at 23:18
  • $\begingroup$ @DonaldSplutterwit So the general expression should have been $\frac{1}{2}(a_n - (a_k-1))(a_k + a_n)$ ? $\endgroup$ – leoybkim Oct 13 at 23:23
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    $\begingroup$ \begin{eqnarray*} \sum_{k=2}^{n}a_k = a_2+ \cdots + a_n = \frac{1}{2} \underbrace{(n-1)}_{ \text{Number of terms} } ( \underbrace{a_2}_{\text{First term}} +\underbrace{ a_n }_{\text{Last term}}). \end{eqnarray*} $\endgroup$ – Donald Splutterwit Oct 13 at 23:27
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    $\begingroup$ Thank you, I understand it now! $\endgroup$ – leoybkim Oct 13 at 23:31
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Part of your confusion is in counting terms.

Note that when you start with $a_2$ and go to $a_n$ you have $n-1$ terms but when you start with ${a_1}$ you have $n$ terms.

The sum is always the number of terms times the average of the first and the last.

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