# arithmetic series general equation

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

• \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*} – Donald Splutterwit Oct 13 at 23:18
• @DonaldSplutterwit So the general expression should have been $\frac{1}{2}(a_n - (a_k-1))(a_k + a_n)$ ? – leoybkim Oct 13 at 23:23
• \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*} – Donald Splutterwit Oct 13 at 23:27
• Thank you, I understand it now! – leoybkim Oct 13 at 23:31

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.