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Visit Summation Sign and Double Summation first if you are not familiar with double summation notation.
-
Identity 1
Double Summation of a Constant Rule.
Explanation and examples
\begin{equation}
\begin{split}
\sum_{{i=\color{red}{n}}}^{\color{green}{k}}\sum_{{j=\color{red}{s}}}^{\color{green}{t}} C &=C\sum_{{i=\color{red}{n}}}^{\color{green}{k}}\sum_{{j=\color{red}{s}}}^{\color{green}{t}}1
\\&=C\cdot (\color{green}{k}-\color{red}{n}+1)(\color{green}{t}-\color{red}{s}+1)
\end{split}
\end{equation}
\begin{equation}
\begin{split}
&\sum_{{i=n}}^{k}\sum_{{j=s}}^{t} C =C\sum_{{i=n}}^{k}\sum_{{j=s}}^{t}1
\\&=C\cdot (k-n+1)(t-s+1)
\end{split}
\end{equation}
Where C is a constant - any expression that
doesn't involve index variables
-
Identity 2
Distributivity.
Explanation and examples
\begin{equation}
\begin{split}
\Big(\sum_{{i=n}}^{k}a_i \Big)\cdot \Big( \sum_{{j=s}}^{t} b_j \Big)= \sum_{{i=n}}^{k}\sum_{{j=s}}^{t} a_i b_j
\end{split}
\end{equation}
\begin{equation}
\begin{split}
&\Big(\sum_{{i=n}}^{k}a_i \Big)\cdot \Big( \sum_{{j=s}}^{t} b_j \Big)=\\&= \sum_{{i=n}}^{k}\sum_{{j=s}}^{t} a_i b_j
\end{split}
\end{equation}
-
Identity 3
Distributivity Special Case.
Explanation with examples
\begin{equation}
\begin{split}
\Big(\sum_{i=1}^{n} x_i\Big)^2 =\Big(\sum_{i=1}^{n} x_i^2\Big)+\sum_{i=1 \\ \quad \color{red}{i\neq j}}^{n}\sum_{j=1}^{n} x_i x_j
\end{split}
\end{equation}
\begin{equation}
\begin{split}
\Big(\sum_{i=1}^{n} x_i\Big)^2 &=\Big(\sum_{i=1}^{n} x_i^2\Big)+\\&+\sum_{i=1 \\ \quad \color{red}{i\neq j}}^{n}\sum_{j=1}^{n} x_i x_j
\end{split}
\end{equation}
-
Identity 4
Distributivity Allows Factorization
Product in Double Summation
\begin{equation}
\begin{split}
\sum_{{i=n}}^{k}\sum_{{j=s}}^{t} a_i b_j=\Big(\sum_{{i=n}}^{k}a_i \Big)\cdot \Big( \sum_{{j=s}}^{t} b_j \Big)
\end{split}
\end{equation}
\begin{equation}
\begin{split}
\sum_{{i=n}}^{k}\sum_{{j=s}}^{t} a_i b_j=\Big(\sum_{{i=n}}^{k}a_i \Big)\cdot \Big( \sum_{{j=s}}^{t} b_j \Big)
\end{split}
\end{equation}
-
Same index in lower-upper limit
How to solve the following expression?
Explanation with examples to practice
\begin{equation}
\begin{split}
\sum_{{\color{red}{i}=k}}^{n}\sum_{{j=k}}^{\color{red}{i}}a_i b_j =?
\end{split}
\end{equation}
Double Summation Rules
The following rules apply to finite sums (both upper and lower limits are integers)
If you are not confident in double summation, check first How
to calculate double summation?
1. Double Summation of a Constant Rule.
\begin{equation}
\begin{split}
\sum_{{i=n}}^{k}\sum_{{j=s}}^{t} C &=C\sum_{{i=n}}^{k}\sum_{{j=s}}^{t}1
\\&=C\cdot (k-n+1)(t-s+1)
\end{split}
\end{equation}
\begin{equation}
\begin{split}
&\sum_{{i=n}}^{k}\sum_{{j=s}}^{t} C =C\sum_{{i=n}}^{k}\sum_{{j=s}}^{t}1
\\&=C\cdot (k-n+1)(t-s+1)
\end{split}
\end{equation}
The rule tells us we can pull the constant C out of the summations signs. Then
the sum is equal to the product of the difference of an upper and lower limit plus one.
Check Double Summation of a Constant.
Let's see how it works with an example:
$\sum_{i=5}^8\sum_{j=2}^7 10=10\sum_{i=5}^8\sum_{j=2}^7=\\
=10\cdot (8-5+1)\cdot (7-2+1)=\\
=10\cdot 4\cdot 6=240$
$\sum_{i=5}^8\sum_{j=2}^7 10=\\10\sum_{i=5}^8\sum_{j=2}^7=\\
=10\cdot (8-5+1)\times \\ \times (7-2+1)=\\
=10\cdot 4\cdot 6=240$
Proof:
$\sum_{{i=n}}^{k}\sum_{{j=s}}^{t} C= $
Let's emphasize the order of operations:
$=\sum_{{i=n}}^{k}\Big(\sum_{{j=s}}^{t} C\Big)= $
Calculating the inner sum (there are t-s+1 of C):
$=\sum_{{i=n}}^{k}C (t-s+1)= $
Again, $C\cdot (t-s+1)$ is just a constant, because there isn't any index variable "i".
So there is $(k-n+1)$ times constant $C(t-s+1).$ So:
$=C\cdot (t-s+1)\times \\ \times (k-n+1) $
Examples to practice:
Use the above rule to solve:
(you can check solutions below)
Example 1 (to practice)
$ \sum_{i=1} ^{10}\sum_{j=1}^7 5$
Example 2 (to practice)
$ \sum_{i=0} ^{5}\sum_{j=1}^{20} p$
Example 3 (to practice)
$ \sum_{i=12} ^{50}\sum_{j=1}^{8} 1$
Example 4 (to practice)
$ \sum_{i=25} ^{35}\sum_{j=22}^{28} k^2$
2. Distributivity:
\begin{equation}
\begin{split}
\Big(\sum_{{i=n}}^{k}a_i \Big)\cdot \Big( \sum_{{j=s}}^{t} b_j \Big)=\sum_{{i=n}}^{k}\sum_{{j=s}}^{t} a_i b_j
\end{split}
\end{equation}
\begin{equation}
\begin{split}
&\Big(\sum_{{i=n}}^{k}a_i \Big)\cdot \Big( \sum_{{j=s}}^{t} b_j \Big)=\\& =\sum_{{i=n}}^{k}\sum_{{j=s}}^{t} a_i b_j
\end{split}
\end{equation}
Where $a_i$ is anything involving "i" (e.g. $2i$, $i^2$, $(i+1)$, member of a set and so on)
- $b_j$ any expression involving index "j"
Explanation Example
What is the difference?
The left side is the product of two summations. The right side tells you do
the inner summation first,
then the outer summation.
Let's show the left-hand side is the same as the right-hand side in following example:
$\color{green}{ \Big(\sum_{{i=3}} ^{4}i\Big) \Big(\sum_{j=5}^{6}j\Big)} \stackrel{?}{=}
\color {red}{\sum_{{i=3}}^{4}\sum_{{j=5}}^{6} i\cdot j} $
$\color{green}{ \Big(\sum_{{i=3}} ^{4}i\Big) \Big(\sum_{j=5}^{6}j\Big)} \stackrel{?}{=} \\
\quad \stackrel{?}{=} \color {red}{\sum_{{i=3}}^{4}\sum_{{j=5}}^{6} i\cdot j} $
Solution:
Let's first write what the LEFT-hand side is:
$\color{green}{ \Big(\sum_{{i=3}} ^{4}i\Big) \Big(\sum_{j=5}^{6}j\Big)}=(3+4)\cdot (5+6) $
$\color{green}{ \Big(\sum_{{i=3}} ^{4}i\Big) \Big(\sum_{j=5}^{6}j\Big)}=\\ \quad=(3+4)\cdot (5+6) $
And the RIGHT-hand side is:
$\color {red}{\sum_{{i=3}}^{4}\sum_{{j=5}}^{6} i\cdot j}=$
| Let's calculate the inner sum first (we add brackets
to highlight the order)
|=$\color {red}{\sum_{{i=3}}^{4}\color{black}{\Big(}\sum_{{j=5}}^{6} i\cdot j}\color{black}{\Big)}=
$|"i" is constant in the inner sum, so we can pull it out of the sum|:=
$=
\color{red}{\sum_{i=3}^4 \color{black}{\Big(}i\cdot \sum_{j=5}^6 j\color{black}{\Big)} } $=|and calculating the inner sum we get:|
$=\color{red}{\sum_{i=3}^4} \Big(\color{red}{i}\cdot(5+6)\Big)$=|and now inserting 3 and 4:|
$=\color{red}{3} (5+6)+\color{red}{4}(5+6)$=|factoring out the common term (5+6)|:
$=(3+4)(5+6)$
OK! Finally, compare the LEFT-hand side and the
RIGHT-hand side. They are the same. So the identity works on that example.
It will work in every case when the limits of the summation are finite. You will always get the same set of pairs
$i\cdot j$. Try to imagine it in general case.
Let's solve together an example using Identity 2:
\begin{equation}
\begin{split}
\Big(\sum_{{i=1}}^{3}i^3\Big)\Big(\sum_{{j=1}}^{2}j\Big)=
\end{split}
\end{equation}
Solution:
Using Identity 1, we can write:
$\Big(\sum_{{i=1}}^{3}i^3\Big)\Big(\sum_{{j=1}}^{2}j\Big)= \sum_{{i=1}}^{3}\sum_{{j=1}}^{2}i^3j=$
$\Big(\sum_{{i=1}}^{3}i^3\Big)\Big(\sum_{{j=1}}^{2}j\Big)=\\= \sum_{{i=1}}^{3}\sum_{{j=1}}^{2}i^3j=$
Solving the inner summation, we get:
$=\sum_{{i=1}}^{3}i^31+i^32 =$
Factoring out $i^3:$
$=\sum_{{i=1}}^{3} i^3(1+2)=\sum_{{i=1}}^{3} i^3\cdot 3$
$=\sum_{{i=1}}^{3} i^3(1+2)=\\= \sum_{{i=1}}^{3} i^3\cdot 3$
Pulling 6 out of the summation sign:
$=3\sum_{{i=1}}^{3} i^3$
Now, calculating the simple single sum :
$=3(1+2^3+3^3)=108$
NOTE
Sometimes you can also see the following equivalent expression without
brackets and a multiplication dot - the last one:
\begin{equation}
\begin{split}
\Big(\sum_{{i=n}}^{k}a_i \Big)\cdot \Big( \sum_{{j=s}}^{t} b_j\Big)&=\sum_{{i=n}}^{k}\sum_{{j=s}}^{t} a_i b_j\\&=
\sum_{{i=n}}^{k}a_i \sum_{{j=s}}^{t} b_j
\end{split}
\end{equation}
\begin{equation}
\begin{split}
&\Big(\sum_{{i=n}}^{k}a_i \Big)\cdot \Big( \sum_{{j=s}}^{t} b_j\Big)=\\&=\sum_{{i=n}}^{k}\sum_{{j=s}}^{t} a_i b_j\\&=
\sum_{{i=n}}^{k}a_i \sum_{{j=s}}^{t} b_j
\end{split}
\end{equation}
In the last expression, $a_i$ was pulled out as a constant
Example on the third summation in NOTE:
Let's solve together:
\begin{equation}
\begin{split}
\sum_{{j=1}}^{3}j^2 \sum_{{i=0}}^{2} i+1=
\end{split}
\end{equation}
1. So, calculating the inner sum, we get:
$\sum_{{j=1}}^{3}j^2\big(1+2+3 \big)\\
= \sum_{{j=1}}^{3}j^2\cdot 6$
$=6\cdot \sum_{j=1}^3 j^2
\\=6(1^2+2^2+3^2)
\\=6\cdot 14=84 $
3. Distributivity, Special Case:
\begin{equation}
\begin{split}
\Big(\sum_{i=1}^{n} x_i\Big)^2 =\Big(\sum_{i=1}^{n} x_i^2\Big)+\sum_{i=1 \\ \quad \color{red}{i\neq j}}^{n}\sum_{j=1}^{n} x_i x_j
\end{split}
\end{equation}
\begin{equation}
\begin{split}
\Big(\sum_{i=1}^{n} x_i\Big)^2 &=\Big(\sum_{i=1}^{n} x_i^2\Big)+\\&+\sum_{i=1 \\ \quad \color{red}{i\neq j}}^{n}\sum_{j=1}^{n} x_i x_j
\end{split}
\end{equation}
Intuition:
Let's rewrite the left-hand side:
$\Big(\sum_{i=1}^{n} x_i\Big)^2=\Big(\sum_{i=1}^{n} x_i\Big)\Big(\sum_{j=1}^{n} x_j\Big)=$
$\Big(\sum_{i=1}^{n} x_i\Big)^2=\\ =\Big(\sum_{i=1}^{n} x_i\Big)\Big(\sum_{j=1}^{n} x_j\Big)$
Using identity 3, we can write
$=\sum_{i=1}^{n} \sum_{j=1}^{n} x_i x_j=$
This double summation represents all possible product pairs $x_i x_j$ where i,j go from 1 to n.
We can rewrite that double sum and split it as $x_i x_i$ where i goes from 1 to n, and
$x_i x_j$ where i and j also go from 1 to n, not including indexes where i and j are the same
as we already wrote those pairs in $x_ix_i$. So:
$=\Big(\sum_{i=1}^{n} x_i^2\Big)+\sum_{i=1 \\ \color{red}{i\neq j}}^{n}\sum_{j=1}^{n} x_i x_j$
$=\Big(\sum_{i=1}^{n} x_i^2\Big)+\\+\sum_{i=1 \\ \color{red}{i\neq j}}^{n}\sum_{j=1}^{n} x_i x_j$
Example:
Express the following double summation using Identity 3 and find the sum:
$\Big(\sum_{i=1}^{n} 2i\Big)^2=$
Solution:
$\Big(\sum_{i=1}^{3} 2i\Big)^2=\Big(\sum_{i=1}^{3} 2i\Big)\Big(\sum_{j=1}^{3} 2i\Big)=$
$\Big(\sum_{i=1}^{3} 2i\Big)^2=\Big(\sum_{i=1}^{3} 2i\Big)\times \\ \times \Big(\sum_{j=1}^{3} 2j\Big)=$
Using Identity 3, we can write:
$=\sum_{i=1}^{3} \sum_{j=1}^{3} 2i\cdot 2j=$
pulling the constant out:
$=4\sum_{i=1}^{3} \sum_{j=1}^{3}ij=$
and finally, using our identity:
$=4\Big(\sum_{i=1}^{3}ii +\sum_{i=1 \\ \hspace{0.1em}\color{red}{i\neq j}}^{3} \sum_{j=1}^{3}ij\Big)$
splitting the sums:
$=4\sum_{i=1}^{3}ii +4\sum_{i=1 \\ \hspace{0.1em}\color{red}{i\neq j}}^{3} \sum_{j=1}^{3}ij$
The first sum is easy:
$=4\sum_{i=1}^{3}ii=4\Big(1 +2^2+3^2\Big)=4\cdot 14=56$
$=4\sum_{i=1}^{3}ii=\\=4\Big(1 +2^2+3^2\Big)=\\=4\cdot 14=56$
In the second sum, we can take the constant "i" out of the inner sum:
$4\sum_{i=1 \\ \hspace{0.1em}\color{red}{i\neq j}}^{3} \sum_{j=1}^{3}ij=
4\sum_{i=1 \\ \hspace{0.1em}\color{red}{i\neq j}}^{3} i\sum_{j=1}^{3}j=$
$4\sum_{i=1 \\ \hspace{0.1em}\color{red}{i\neq j}}^{3} \sum_{j=1}^{3}ij=\\=
4\sum_{i=1 \\ \hspace{0.1em}\color{red}{i\neq j}}^{3} i\sum_{j=1}^{3}j=$
and calculate as we did in the NOTE
when we say "i" is 1, then j can be 2 and 3
but can't be 1, and so on:
$=4(1\cdot(2+3)+2\cdot(1+3)+3\cdot(1+2))=\\
=4\cdot 22=88 $
$=4(1\cdot(2+3)+\\+2\cdot(1+3)+3\cdot(1+2))\\
=4\cdot 22=88 $
finally, summing results:
$= 56+88=144$
Final solution:
$\Big(\sum_{i=1}^{3} 2i\Big)^2=4\sum_{i=1}^{3} i^2+4\sum_{i=1
\\ \hspace{0.1em}\color{red}{i\neq j}}^{3} \sum_{j=1}^{3}ij=\\
=56+88=144$
$\Big(\sum_{i=1}^{3} 2i\Big)^2=4\sum_{i=1}^{3} i^2+\\+4\sum_{i=1
\\ \hspace{0.1em}\color{red}{i\neq j}}^{3} \sum_{j=1}^{3}ij=\\
=56+88=144$
Note: to calculate the above sum, it is easier to use Identity 2
Example 5 (to practice)
$\Big( \sum_{i=1} ^4 i \Big)^2=$
4. Distributivity Allows Factorization:
The previous formula reversed. From Identity 1, it is obvious that
the following reversed identity will also work:
\begin{equation}
\begin{split}
\sum_{{i=n}}^{k}\sum_{{j=s}}^{t} a_i b_j=\Big(\sum_{{i=n}}^{k}a_i \Big)\cdot \Big( \sum_{{j=s}}^{t} b_j \Big)
\end{split}
\end{equation}
\begin{equation}
\begin{split}
\sum_{{i=n}}^{k}\sum_{{j=s}}^{t} a_i b_j&=\Big(\sum_{{i=n}}^{k}a_i \Big)\times \\&\times \Big( \sum_{{j=s}}^{t} b_j \Big)
\end{split}
\end{equation}
Let's solve together an example using Identity 4:
\begin{equation}
\begin{split}
\sum_{{j=7}}^{9}\sum_{{i=1}}^{3} ji=
\end{split}
\end{equation}
Solution:
Using Identity 2, we can write:
$\sum_{{j=7}}^{9}\sum_{{i=1}}^{3} ji= \Big(\sum_{{j=7}}^{9}j\Big)\Big(\sum_{{i=1}}^{3}i\Big)=$
$\sum_{{j=7}}^{9}\sum_{{i=1}}^{3} ji=\\= \Big(\sum_{{j=7}}^{9}j\Big)\Big(\sum_{{i=1}}^{3}i\Big)$
Calculating each of the sums on the right, we get:
$=(7+8+9)(1+2+3)=24\cdot 6=144$
$=(7+8+9)(1+2+3)\\=24\cdot 6=144$
Let's solve another one together, and then you will do a couple on your own:
\begin{equation}
\begin{split}
\sum_{{i=3}}^{5}\sum_{{j=20}}^{21} i^i j^2=
\end{split}
\end{equation}
Solution:
Using Identity 2, we can write:
$ \sum_{{i=3}}^{5}\sum_{{j=20}}^{21} i^i j^2=
(\sum_{{i=3}}^{5} i^i )(\sum_{{j=20}}^{21}j^2)=$
$ \sum_{{i=3}}^{5}\sum_{{j=20}}^{21} i^i j^2= \\
\\=(\sum_{{i=3}}^{5} i^i )(\sum_{{j=20}}^{21}j^2)$
Calculating the sums and multiplying, we get:
$=(3^3+4^4+5^5)(20^2+21^2)=3408\cdot 841=$
$=(3^3+4^4+5^5)(20^2+21^2)\\=3408\cdot 841=$
$=2866128$
Now, it is your turn!
Solve the following
double summation examples using Identity 2
(you can check solutions below):
Example 6 (to practice)
$ \sum_{i=1} ^{2}\sum_{j=1}^3{i^2j^2}$
Example 7 (to practice)
$ \sum_{i=1} ^{3}\sum_{j=1}^2\frac{i}{3}(j+1)$
Example 8 (to practice)
$ \sum_{i=10} ^{12}\sum_{j=11}^ {13} 5ij$
Example 9 (to practice)
$ \sum_{i=2} ^{3}\sum_{j=2} ^{3}i^ij^j$
Example 10 (to practice)
$ \sum_{i=8} ^{10}\sum_{j=11} ^{12}(i+3)(j-1)$
Example 10 (to practice)
$ \sum_{i=8} ^{10}\sum_{j=11} ^{12} (i+3)\times \\ \times (j-1)$
Same Index in Both Summation Signs Limits:
How to solve an expression like that?
\begin{equation}
\begin{split}
\sum_{{\color{red}{i}=k}}^{n}\sum_{{j=k}}^{\color{red}{i}}a_i b_j =?
\end{split}
\end{equation}
It will be easier to show step by step with an example:
Example
$\sum_{{\color{red}{i}=1}}^{3}\sum_{{j=1}}^{\color{red}{i}}ij = $
The lower limits are the same and are equal to 1.
STEPS:
STEP 1: we start from the outer sum - i is equal to 1.
Inner sum goes from 1 to 1, also we plug in 1 for i in ij:
outer i=1
inner sum:
$\sum_{{j=1}}^{\color{red}{1}}1\cdot j=1\cdot 1=1$
outer i=2
inner sum:
$\sum_{{j=1}}^{\color{red}{2}}2\cdot j=2\cdot 1+2\cdot 2=6$
$\sum_{{j=1}}^{\color{red}{2}}2\cdot j=\\=2\cdot 1+2\cdot 2=6$
and finally, outer i=3
inner sum:
$\sum_{{j=1}}^{\color{red}{3}}3\cdot j=3\cdot 1+3\cdot 2+\\
+3\cdot 3=18$
$\sum_{{j=1}}^{\color{red}{3}}3\cdot j=3\cdot 1+\\+3\cdot 2
+3\cdot 3=18$
STEP 2: sum the results from each of outer i:
$=1+6+18=25$ the end
Final solution in one step:
\begin{equation}
\begin{split}
\sum_{{\color{red}{i}=1}}^{3}\sum_{{j=1}}^{\color{red}{i}}ij=&
\sum_{{j=1}}^{\color{red}{1}}\color{red}{1}\cdot j+ \sum_{{j=1}}^{\color{red}{2}}\color{red}{2}\cdot j+
\\&+\sum_{{j=1}}^{\color{red}{3}}\color{red}{3}\cdot j \\&=1+6+18 \\&=25
\end{split}
\end{equation}
\begin{equation}
\begin{split}
&\sum_{{\color{red}{i}=1}}^{3}\sum_{{j=1}}^{\color{red}{i}}ij=\\&
=\sum_{{j=1}}^{\color{red}{1}}\color{red}{1}\cdot j+ \sum_{{j=1}}^{\color{red}{2}}\color{red}{2}\cdot j +
\\&+\sum_{{j=1}}^{\color{red}{3}}\color{red}{3}\cdot j= \\&=1+6+18 \\&=25
\end{split}
\end{equation}
Let's solve together:
\begin{equation}
\begin{split}
\sum_{{\color{red}{i}=1}}^{3}\sum_{{j=1}}^{\color{red}{i}}i-j =
\end{split}
\end{equation}
Following the above solution, we will plug in in the upper limit ofthe inner
sum
numbers from 1 to 3 (as outer "i" goes from 1 to 3)
and then add the sums:
$ = \sum_{{j=1}}^{\color{red}{1}}1-j+\sum_{{j=1}}^{\color{red}{2}}2-j+\\
+\sum_{{j=1}}^{\color{red}{3}}3-j=
$
$ = \sum_{{j=1}}^{\color{red}{1}}1-j+\\+\sum_{{j=1}}^{\color{red}{2}}2-j+\\
+\sum_{{j=1}}^{\color{red}{3}}3-j=
$
Now, calculating each of the sums:
$=(1-1)+(2-1+2-2)+\\+(3-1+3-2+3-3)=4$
$=(1-1)+\\+(2-1+2-2)+\\+(3-1+3-2+3-3)\\=4$
Now it is your turn!
Solve the following
double summation examples with the same lower-upper index.
(you can check solutions below):
Example 11 (to practice)
$ \sum_{i=0}^{2}\sum_{j=0}^{i} 2i-3j$
Example 12 (to practice)
$ \sum_{i=1}^{3}\sum_{j=1}^{i} i^2j$
Example 13 (to practice)
$ \sum_{i=1}^{4}\sum_{j=1}^{i} i+j$
Solutions are at the bottom on purpose. First, calculate yourself :)) and then scroll down.
Solutions
1. 350
2. 120p
3. 312
4. $77k^2$
5. 42p
6. 70
7. 10
8. 5940
9. 961
10. 756
11. 4
12. 67
13. 50