Simple Double Summation I understand how to sum a single sum, but I don't know how to solve a double sum without explicit limits.  Please help guide me in the right direction to solve problems 3 through 5 in the included image.  Thank you!!

  
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*$a_n = \sum\limits_{i=1}^n (2i-1)$
  
*$a_n = \sum\limits_{i=1}^n (3i^2-3i+1)$
  
*$a_n= \sum\limits_{i=1}^n \sum\limits_{j=1}^n 1$
  
*$a_n= \sum\limits_{i=1}^n \sum\limits_{j=1}^n i$
  
*$a_n= \sum\limits_{i=1}^n \sum\limits_{j=1}^n j$
  


 A: Note: The first (and original) part of this answers solves a harder problem than was actually asked, but if you’re taking a discrete math course, you’ll probably be doing similar things before too long.
I’ll do (d) and leave you with a couple of hints for (c) and (e).
In order to evaluate $\sum_{i=1}^n\sum_{j=1}^ii$, your first step should be to evaluate the inner sum, $\sum_{j=1}^ii$. Since $j$, the index of summation, runs from $1$ through $i$, this is a sum of $i$ terms. And since $i$, the general term in the summation, does not depend on $j$, this inner summation is just
$$\sum_{j=1}^ii=\underbrace{i+i+\ldots+i}_{i}=i^2\;.$$
The original summation can now be reduced to a single summation:
$$\sum_{i=1}^n\sum_{j=1}^ii=\sum_{i=1}^ni^2\;;$$
it’s the sum of the first $n$ squares. This is a formula that you will probably be expected to learn:
$$\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6\;,\tag{1}$$
which is the final answer for (d).
You can do (c) and (e) in the same way: start by evaluating the inner sum. For (c), how many copies of $1$ are you adding? For both (c) and (d) you will need to know that
$$\sum_{i=1}^ni=\frac{n(n+1)}2\;;$$
this is an extremely useful formula that you will certainly be expected to learn, if you don’t already know it. For (e) you’ll need formula $(1)$ as well. Finally, don’t forget that you can always pull a constant factor out of a sum:
$$\sum_{i=1}^nca_i=c\sum_{i=1}^na_i\;.$$
Added: I’ve actually done more here than was called for: I’ve shown you how to write a general formula for the sum, so that you can just plug in $n=1,2,3,4,5$. Of course you don’t need that just to get the first five terms. Here, for instance, is a brute force calculation of the third term of (c):
$$\begin{align*}
\sum_{i=1}^3\sum_{j=1}^ii&=1+(1+2)+(1+2+3)\\
&=1+3+6\\
&=10\;.
\end{align*}$$
The others are equally straightforward.
A: $$\sum_{i=1}^n{\sum_{j=1}^n{1}}=\sum_{i=1}^nn=n\sum_{i=1}^n1=n\cdot{n}=n^2$$
Use this to solve 5.
A: Hint 1: $\sum_{i=1}^2 \sum_{j=1}^3 1 = \sum_{i=1}^2 (\sum_{j=1}^3 1) = \sum_{i=1}^2 ( 1 + 1 + 1) = \sum_{i=1}^2 3 = 3 + 3 = 6$.
Hint 2: $\sum_{i=1}^2 \sum_{j=1}^3 i = \sum_{i=1}^2 (\sum_{j=1}^3 i) = \sum_{i=1}^2 ( i + i + i) = \sum_{i=1}^2 (3i) =  3\sum_{i=1}^2 i = 3(1+2) = 9$.
Hint 3: $\sum_{i=1}^2 \sum_{j=1}^3 j = \sum_{i=1}^2 (\sum_{j=1}^3 j) = \sum_{i=1}^2 ( 1 + 2 + 3) = \sum_{i=1}^2 6 =  6+6 = 12$.
A: It's a simple matter of thinking recursively. How does one do a sum, say $f(i)$, for some function $f$? For each $i$ you calculate the value of $f(i)$ and then sum them. In this case, say for problem $(d)$, the function you are summing is again a sum, so for each $i$ we must do a whole sum.
Let's calculate $a_{3}$ for $(d)$. 
$a_{3} = \sum_{i=1}^{3} \sum_{j=1}^{i} i$.
Now ranging over the $i$'s, this is
$a_{3} = (\sum_{j=1}^{1} 1) + (\sum_{j=1}^{2} 2) + (\sum_{j=1}^{3} 3)
= 1 + 4 + 9 =14.$
