2
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy