Double Summation Question Given ${\displaystyle \; \sum _{i=1}^{n}\sum _{j=i}^{i+1}(3i+j)}$ change it to form ${\displaystyle \; \sum _{j}^{}\sum _{i}^{}}(3i+j)$ and calculate.
Would it be valid to subtract $i$ and then add $1$ to ${\displaystyle \; \sum _{j=i}^{i+1}(3i+j)}$ so it's  ${\displaystyle \; \sum _{j=1}^{2}}(3(i+i-1)+j)$ and then just do ${\displaystyle \; \sum _{j=1}^{2}\sum _{i=1}^{n}(3(2i-1)+j)}$ ???
 A: What you have to do is a change of summation. The expression $3i+j$ doesn't play any role and as well could be an arbitrary function $\varphi(i,j)$. Notice that as summation is commutative it doesn't matter the order in which you compute the sum. That is, the only relevant issue are the pairs $(i,j)$
whose values $\varphi(i,j)$ you're adding up.
The notation ($\ast$) $\sum_{i=1}^n\sum_{j=i}^{i+1}\varphi(i,j)$ is telling you to add
the numbers $\varphi(i,j)$ with the order $\varphi(1,1)+\varphi(1,2)+\varphi(2,2)+\varphi(2,3)+\cdots+\varphi(n,n)+\varphi(n,n+1)$. Again, you can add this in the order you like, so the important issue is over what pairs you're adding up. This being said, what you're being asked is to express this sum in such an order that it looks like $\sum_j\sum_i\varphi(i,j)$. This can be a little confusing sometimes so I always like to plot the pairs $(i,j)$
I'm adding up in a integer plane. In your case (and with $n=6$) it would look like

The green and red arrows is what I would paint next. The red ones describe the order of summation in the first expression ($\ast$); first you fix $i$ and then you add the corresponding values when $j=i$ and $j=i+1$. Now, the form you're being asked to express it in corresponds to the green arrows; first fix $j$ and then add the corresponding values corresponding to each $i$. Except for the points $(1,1)$ and $(6,7)$, you can see that for every $j$ you have to add the corresponding values when $i=j-1$ and $i=j$. That is
$$
\sum_{i=1}^6\sum_{j=i}^{i+1}\varphi(i,j)=\varphi(1,1)+\varphi(6,7)+\sum_{j=2}^6\sum_{i=j-1}^j\varphi(i,j).
$$
Sure you can see that there is nothing special with $n=6$, and that the general answer is
$$
\sum_{i=1}^n\sum_{j=i}^{i+1}\varphi(i,j)=\varphi(1,1)+\varphi(n,n+1)+\sum_{j=2}^n\sum_{i=j-1}^j\varphi(i,j).
$$
Finally, if you're asked to express this sum exactly in the form $\sum_j\sum_i\varphi(i,j)$, then you can simply define $\varphi(0,1)$ and $\varphi(n+1,n+1)$ to be $0$ and then express it like $\sum_{j=1}^{n+1}\sum_{i=j-1}^j\varphi(i,j)$. Hope this clarify any future confussion about changing the order in summation!
(Edit; just seen that is asked to find the value of the sum) Setting $\varphi(i,j)=3i+j$ in the sum after the change of summation you get the value
$$
4+3n+n+1+\sum_{j=2}^n\sum_{i=j-1}^j3i+j=5+4n+\sum_{j=2}^n[3(j-1)+j+3j+j]=5+4n-3(n-1)+8\sum_{j=2}^nj=8+n+8(\frac{n(n+1)}{2}-1)=\cdots,
$$
where I've used that $\sum_{j=1}^nj=\frac{n(n+1)}{2}$, which is a well-known identity.
