How to Prove that $\frac{S_i}{i}(j-k) + \frac{S_j}{j}(k-i) +\ \frac{S_k}{k}(i-j)$ is independent of $i,j,k$ Let $S_i$ be the sum of the first $i$ terms of the arithmetic sequence $a_1,a_2,a_3\ldots $. Show that the value of the expression
$$\frac{S_i}{i}(j-k) + \frac{S_j}{j}(k-i) +\ \frac{S_k}{k}(i-j)$$
does not depend on the numbers $i,j,k$ nor on the choice of the arithmetic sequence $a_1,a_2,a_3,\ldots$
Any nice approach to this ? 
 A: Let $a_n=a_1+(n-1)d$. So we have $S_n=\frac{n}{2}(a_1+a_n)=\frac{n}{2}(2a_0+nd)$.
Hence,
\begin{align}
\text{LHS}&=\frac{S_i}{i}(j-k) + \frac{S_j}{j}(k-i) +\ \frac{S_k}{k}(i-j)\\&= 
\frac{\frac{i}{2}(2a_0+id)}{i}(j-k) + \frac{\frac{j}{2}(2a_0+jd)}{j}(k-i) +\ \frac{\frac{k}{2}(2a_0+kd)}{k}(i-j)\\
&=\frac{(2a_0+id)(j-k)+(2a_0+jd)(k-i)+(2a_0+kd)(i-j)}{2}\\
&=a_1\overbrace{(j-k+k-t+i-j)}^0+\frac{d}{2}\overbrace{\left(ij-ik+jk-ij+ik-jk\right)}^0\\
&=0
\end{align}
A: Note that the sum of the first $n$ terms of an arithmetic progression is given by: $$S_n =n\frac{[2a +(n-1)d]}{2} \implies \frac{S_n}{n}=\frac{[2a +(n-1)d]}2$$
Now, we require, $$\frac{S_i}{i}(j-k) + \frac{S_j}{j}(k-i) + \frac{S_k}{k}(i-j)$$ $$=\frac12([2a](i-j+j-k+k-i)+d[(i-1)(j-k)+(j-1)(k-i)+(k-1)(i-j)])$$ $$= \, ?$$
A: If $\{a_n\}_{n=1}^\infty$ is an arithmetic sequence
$(a_\ell = a_1 + (\ell-1)d)$, then 
$\qquad (1.)\quad S_\ell = \sum_{n=1}^\ell a_n = \dfrac{\ell}{2}(a_1+a_\ell)$
$\qquad (2.)\quad a_u-a_v=(u-v)d$
So
\begin{align}
   \frac{S_i}{i}(j-k) + \frac{S_j}{j}(k-i) +\ \frac{S_k}{k}(i-j) 
   &= \frac 12(a_1+a_i)(j-k) + \frac 12(a_1+a_j)(k-i) + \frac 12(a_1+a_k)(i-j)\\
   &= \frac i2(a_k-a_j)+\frac j2(a_i-a_k)+\frac k2(a_j-a_i) \\
   &= \frac {id}2(k-j)+\frac{jd}2(i-k)+\frac{kd}2(j-i) \\
   &= 0
\end{align}
A: For a given AP, the sum of first $n$ terms is of the form $An^2+Bn$ for some constants $A,B$
Hence $\dfrac{S_i}{i} = \dfrac{Ai^2+Bi}{i} = Ai+B$
Hence the given sum is simply $\displaystyle \sum_{cyc} (Ai+B)(j-k)$
$=\displaystyle A \sum_{cyc} i(j-k) +\displaystyle B \sum_{cyc} (j-k) =0 $
