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!!

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

double sum homework

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    $\begingroup$ Note that your answers for the single sums, a and b, are incorrect. You seem to be confusing the $n$th term with the $n$th sum. $\endgroup$ – Jonas Meyer Jun 28 '13 at 20:25
  • $\begingroup$ Yes, J@JonasMeyer is correct. The idea is to determine the first five $a_n$, not the first $5$ terms in the sum. $\endgroup$ – Thomas Andrews Jun 28 '13 at 20:27
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    $\begingroup$ Oh! Thanks for that. So a) would be: 1, 4, 9, 16, 25 ? $\endgroup$ – user84059 Jun 28 '13 at 20:30
  • $\begingroup$ @DiscreteMath: Yes! $\endgroup$ – Jonas Meyer Jun 28 '13 at 20:33
  • $\begingroup$ @DiscreteMath Yes, that is the answer for (a). $\endgroup$ – Thomas Andrews Jun 28 '13 at 20:33

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


The original summation can now be reduced to a single summation:


it’s the sum of the first $n$ squares. This is a formula that you will probably be expected to learn:


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


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:


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.

  • $\begingroup$ Thank you. I have seen that formula before (sum of squares). I actually just saw that in my book. However, I don't understand why i+i+i+i+i = i^2. Can you elaborate on that point please? $\endgroup$ – user84059 Jun 28 '13 at 20:27
  • $\begingroup$ (Deleted obsolete comment) The general formulas are great for generalizing, but of course they are not needed to find the first 5 sums in any of these problems. $\endgroup$ – Jonas Meyer Jun 28 '13 at 20:27
  • $\begingroup$ @DiscreteMath: It’s not that $i+i+i+i+i=i^2$; that’s true only if $i=5$. What’s true is that $$\underbrace{i+i+\ldots+i}_{i\text{ copies of }i}=i^2\;.$$ If you add up $j$ copies of $i$, you get $ji$; when $j=i$, that’s $i^2$. $\endgroup$ – Brian M. Scott Jun 28 '13 at 20:31
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    $\begingroup$ @Jonas: Yes, I got a bit ahead of myself (and, in all likelihood, the course). $\endgroup$ – Brian M. Scott Jun 28 '13 at 20:32
  • $\begingroup$ @Brian: So you're stating that 1 + 2 + 3 + 4 + 5 from j = 1 to j = i (i=5) is i^2? But 1+2+3+4+5 = 15, right? This is where I'm lost.. $\endgroup$ – user84059 Jun 28 '13 at 20:40

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: $\sum_{i=1}^2 \sum_{j=1}^3 j = \sum_{i=1}^2 (\sum_{j=1}^3 j) = \sum_{i=1}^2 ( 1 + 2 + 3) = \sum_{i=1}^2 6 = 6+6 = 12$.


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

$a_{3} = (\sum_{j=1}^{1} 1) + (\sum_{j=1}^{2} 2) + (\sum_{j=1}^{3} 3) = 1 + 4 + 9 =14.$


$$\sum_{i=1}^n{\sum_{j=1}^n{1}}=\sum_{i=1}^nn=n\sum_{i=1}^n1=n\cdot{n}=n^2$$ Use this to solve 5.

  • $\begingroup$ You may freely exchange the order of summation for finite sums, however. $\endgroup$ – Austin Mohr Jun 28 '13 at 20:14
  • $\begingroup$ yes, this is correct...i was just trying to make it easy to see initially that you do the right as a single summation, solve, and then do the remaining. $\endgroup$ – Eleven-Eleven Jun 28 '13 at 20:17

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