Compute $S_1=a_1a_2+a_2a_3+...+a_{n-1}a_n$ and $S_2$, where $a_1,a_2,...,a_n$ is an arithmetic progression

Question

Given $(a_n)$ is an arithmetic progression, where the difference between two consecutive terms is $r$, compute, relative to $a_1$ and $r$:

\begin{equation}S_1=a_1a_2+a_2a_3+...+a_{n-1}a_n\end{equation}

and

\begin{equation}S_2=\frac{1}{a_1a_2a_3}+\frac{1}{a_2a_3a_4}+\frac{1}{a_{n-2}a_{n-1}a_n}\end{equation}

I tried

\begin{equation}S_1=\Sigma{a_ka_{k+1}=\Sigma{a_k(a_k+r)}=\Sigma{a_k^2}+r\Sigma{a_k}}\end{equation}

where

\begin{equation}\sum\limits_{k=1}^{n-1} a_k=\frac{(a_1+a_{n-1})(n-1)}{2}\end{equation}

and

\begin{equation}a_{n-1}=a_1+(n-2)r\end{equation}

but I can not manage to compute $\Sigma{a_k^2}$

Answer 1

Extend definition of $a_k$ to all integers by $a_k = a_1 + (k-1)r$ for $k \le 0$ or $\ge n$, we have $$\begin{align} a_k a_{k+1} &= \frac{a_k a_{k+1} a_{k+2} - a_{k-1} a_{k} a_{k+1}}{a_{k+2}-a_{k-1}} = \frac{1}{3r}(a_k a_{k+1} a_{k+2} - a_{k-1} a_{k} a_{k+1})\\ \frac{1}{a_k a_{k+1} a_{k+2}} &= \frac{1}{a_{k+2}-a_k}\left(\frac{a_{k+2} - a_k}{a_{k}a_{k+1}a_{k+2}}\right) = \frac{1}{2r}\left(\frac{1}{a_k a_{k+1}} - \frac{1}{a_{k+1}a_{k+2}}\right) \end{align}$$

Both $S_1$ and $S_2$ are telescoping sums. We have:

$$\begin{align} S_1 &= \frac{1}{3r}(a_{n-1}a_{n}a_{n+1} - a_0a_1a_2)\\ &= \frac{1}{3r}\left((a_1+(n-2)r)(a_1+(n-1)r)(a_1+nr) - (a_1-r)a_1(a_1+r)\right)\\ \\ S_2 &= \frac{1}{2r}\left(\frac{1}{a_1a_2} - \frac{1}{a_{n-1}a_n}\right)\\ &= \frac{1}{2r}\left(\frac{1}{a_1(a_1+r)} - \frac{1}{(a_1+(n-2)r)(a_1+(n-1)r)}\right) \end{align} $$

Answer 2

$$\sum_{i=1}^na_i^2=\sum_{i=1}^n(a_1+(i-1)r)^2=\sum_{i=1}^na_1^2+2ra_1(i-1)+r^2(i-1)^2=na_1^2+2ra_1\sum_{i=1}^n(i-1)+r^2\sum_{i=1}^n(i-1)^2=na_1^2+\frac{2ra_1n(n-1)}{2}+\frac{r^2n(n-1)(2n-1)}{6}$$