• How to calculate double summation?
• Product in double summation
• One variable in double summation
• Double summation of a constant

How to calculate double summation?

\[ \text{$\sum_{\color{blue}{i=1}} ^{2} \sum_{\color{red}{j=1}} ^{3}{\color{blue}{\color{blue}{i}+\color{red}{j}}} $}\]

Read in Ukrainian

\begin{equation} \begin{split} \sum_{\color{blue}{i=1}}^{2} \sum_{\color{red}{j=1}} ^{3}{ \color{blue}{i}+\color{red}{j}} &= \sum_{\color{blue}{i=1}}^{2}(\color{blue}{i}+\color{red}{1})+\\&+ (\color{blue}{i}+\color{red}{2})+\\&+(\color{blue}{i}+\color{red}{3})=\\ only\hspace{0.4em} \color{red}{j}\hspace{0.4em} was \\ \hspace{0.4em} changing..!\\ blue\\\hspace{0.4em}remained\hspace{0.4em}\\ constant\\ & =\sum_{\color{blue}{i=1}}^{2} (3\cdot \color{blue}{i}+\color{red}{6}) \end{split} \end{equation}

\begin{equation} \begin{split} \sum_{\color{blue}{i=1}}^{2} \sum_{\color{red}{j=1}} ^{3}{ \color{blue}{i}+\color{red}{j}} &= \sum_{\color{blue}{i=1}}^{2}(\color{blue}{i}+\color{red}{1})+(\color{blue}{i}+\color{red}{2})+(\color{blue}{i}+\color{red}{3})=\\ \ only\hspace{0.4em} \color{red}{j}\hspace{0.4em} was \\ \hspace{0.4em} changing..!\\ blue\\\hspace{0.4em}remained\hspace{0.4em}\\ constant\\ & =\sum_{\color{blue}{i=1}}^{2} (3\cdot \color{blue}{i}+\color{red}{6}) \end{split} \end{equation}

\begin{equation} \begin{split} &\sum_{\color{blue}{i=1}}^{2} (3\cdot \color{blue}{i}+6)=\\&=(3\cdot \color{blue}{1}+6)+\\&+(3\cdot \color{blue}{2}+6)=\\& =9+12=21 \end{split} \end{equation}

\begin{equation} \begin{split} \sum_{\color{blue}{i=1}}^{2} (3\cdot \color{blue}{i}+6)&=(3\cdot \color{blue}{1}+6)+(3\cdot \color{blue}{2}+6)=\\& =9+12=21 \end{split} \end{equation}

Not that scary, right?
Now you can try to calculate the following example yourself! You can check the solution as usual at the bottom of this page.

The example above is very similar to the one I explained, so you shouldn't have any problems while following STEP 1 and STEP 2.

What MISTAKE can you probably make during calculation? This one

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Now let's do more examples together:

Remember, the summation index can be any letter; i and j are just the most popular ones.
We will go against the world and use l and k in the following example so you will get used to see different letters. Be happy I didn't choose $\xi$ (ksi) and $\eta$ (eta) from the Greek alphabet.

Example

\begin{equation} \begin{split} \sum_{{k=2}}^{4}\sum_{{l=1}}^{2} (2\cdot k-l)= \end{split} \end{equation}

Solution: We will do it just as in the previous two examples, following the steps written at the top of the page.

So first inner sum:

\begin{equation} \begin{split} \sum_{{k=2}}^{4}\sum_{{l=1}}^{2} (2\cdot k-l)&= \sum_{k=2} ^{4}(2\cdot k -1)+(2\cdot k-2)\\&= \sum_{k=2} ^{4}(4\cdot k-3) \\now \hspace{0.4em} outer\hspace{0.4em}\\ sum\hspace{0.4em} as\hspace{0.4em}\\ usual: \\&=(4\cdot 2-3)+(4\cdot 3 -3)+ \\&+(4\cdot 4-3) \\and\hspace{0.4em}after\hspace{0.4em} \\adding\hspace{0.4em} we\\ \hspace{0.4em} get: \\&=27 \end{split} \end{equation}

\begin{equation} \begin{split} &\sum_{{k=2}}^{4}\sum_{{l=1}}^{2} (2\cdot k-l)\\&= \sum_{k=2} ^{4}(2\cdot k -1)+(2\cdot k-2)\\&= \sum_{k=2} ^{4}(4\cdot k-3) \\&=(4\cdot 2-3)+(4\cdot 3 -3)+ \\&+(4\cdot 4-3) \\&=27 \end{split} \end{equation}

Now it's your turn. Solve the following two examples. You have the knowledge needed for it. You can check the solutions at the bottom of the page.

Product in the double summation

Let's solve together:

Find the summation of $xy$

First, we calculate the inner sum, plugging in values from 1 to 2 into variable y. After that, we calculate the single sum with the index variable x:

\begin{equation} \begin{split} \sum_{{x=1}}^{2}\sum_{{y=1}}^{2} (x\cdot y)&=\sum_{{x=1}}^{2}(x\cdot 1)+(x\cdot 2) \\&=(1\cdot 1 + 1\cdot 2) + (2\cdot 1 + 2\cdot 2) \\&=1+2+2+4 \\&=9 \end{split} \end{equation}

\begin{equation} \begin{split} &\sum_{{x=1}}^{2}\sum_{{y=1}}^{2} (x\cdot y)=\\&=\sum_{{x=1}}^{2}(x\cdot 1)+(x\cdot 2) \\&=(1\cdot 1 + 1\cdot 2) +\\&+ (2\cdot 1 + 2\cdot 2) \\&=1+2+2+4 \\&=9 \end{split} \end{equation}

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Now let's consider another similar double summation example with division:

Let's solve together:

\begin{equation} \begin{split} \sum_{{i=5}}^{6}\sum_{{j=1}}^{2} \frac{j^2}{i}&=\sum_{{i=5}}^{6} \frac{1^2}{i} +\frac{2^2}{i} \\&=\Big(\frac{1^2}{5}+\frac{2^2}{5}\Big) +\Big(\frac{1^2}{6}+\frac{2^2}{6}\Big) \\&=\Big(\frac{1+4}{5}\Big)+\Big(\frac{1+4}{6}\Big) \\&=1+\frac{5}{6}=\frac{11}{6} \end{split} \end{equation}

\begin{equation} \begin{split} &\sum_{{i=5}}^{6}\sum_{{j=1}}^{2} \frac{j^2}{i}=\\&=\sum_{{i=5}}^{6} \frac{1^2}{i} +\frac{2^2}{i} \\&=\Big(\frac{1^2}{5}+\frac{2^2}{5}\Big) +\Big(\frac{1^2}{6}+\frac{2^2}{6}\Big) \\&=\Big(\frac{1+4}{5}\Big)+\Big(\frac{1+4}{6}\Big) \\&=1+\frac{5}{6}=\frac{11}{6} \end{split} \end{equation}

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And another one, there is never enough!

Let's solve together:

\begin{equation} \begin{split} \sum_{{i=7}}^{8}\sum_{{j=3}}^{4} \frac{j+1}{(i-1)^2}&=\sum_{{i=7}}^{8} \frac{3+1}{(i-1)^2} + \frac{4+1}{(i-1)^2} \\&= \Big(\frac{3+1}{(7-1)^2}+ \frac{4+1}{(7-1)^2}\Big)+ \\&+\Big(\frac{3+1}{(8-1)^2}+ \frac{4+1}{(8-1)^2}\Big) \\&=\Big(\frac{4}{36}+\frac{5}{36}\Big) +\Big(\frac{4}{49}+\frac{5}{49}\Big) \\&=\frac{3}{12}+\frac{9}{49} \\&=\frac{255}{588}=\frac{85}{196} \end{split} \end{equation}

\begin{equation} \begin{split} &\sum_{{i=7}}^{8}\sum_{{j=3}}^{4} \frac{j+1}{(i-1)^2}=\\&=\sum_{{i=7}}^{8} \frac{3+1}{(i-1)^2} + \frac{4+1}{(i-1)^2} \\&= \Big(\frac{3+1}{(7-1)^2}+ \frac{4+1}{(7-1)^2}\Big)+ \\&+\Big(\frac{3+1}{(8-1)^2}+ \frac{4+1}{(8-1)^2}\Big) \\&=\Big(\frac{4}{36}+\frac{5}{36}\Big) +\Big(\frac{4}{49}+\frac{5}{49}\Big) \\&=\frac{3}{12}+\frac{9}{49} \\&=\frac{255}{588}=\frac{85}{196} \end{split} \end{equation}

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Cool looking example with j to the power of i (easy as all of them):

Let's solve together:

\begin{equation} \begin{split} \sum_{{i=3}}^{4}\sum_{{j=1}}^{2} \frac{i^2}{1+j^i}&= \sum_{{i=3}}^{4} \frac{i^2}{1+1^i} + \frac{i^2}{1+2^i}\\common\hspace{0.4em}\\mistake \\& \color{red} {\neq \frac{3^2}{1+1^3}+\frac{4^2}{1+1^4}}\\ plug\hspace{0.4em}in\hspace{0.4em}3\hspace{0.4em}and\hspace{0.4em}4\\ \hspace{0.4em}in\hspace{0.4em}each\hspace{0.4em}of\hspace{0.4em}2\\ \hspace{0.4em}fractions: \\&= \frac{\color{red}{3^2}}{1+1^\color{red}{3}}+ \frac{\color{red}{3^2}}{1+2^\color{red}{3}}+ \\&+\frac{\color{red}{4^2}}{1+1^\color{red}{4}}+ \frac{\color{red}{4^2}}{1+2^\color{red}{4}} \\&=\frac{9}{2}+\frac{9}{9}+\frac{16}{2}+\frac{16}{17} \\&=\frac{1377+306+2448+288}{306} \\&=\frac{4419}{306}\\divide\hspace{0.4em}each\hspace{0.4em} by\hspace{0.4em}9 \\&=\frac{491}{34} \end{split} \end{equation}

\begin{equation} \begin{split} &\sum_{{i=3}}^{4}\sum_{{j=1}}^{2} \frac{i^2}{1+j^i}\\&= \sum_{{i=3}}^{4} \frac{i^2}{1+1^i} + \frac{i^2}{1+2^i} \end{split} \end{equation}

Common mistake:

\begin{equation} \begin{split} & \color{red} {\neq \frac{3^2}{1+1^3}+\frac{4^2}{1+1^4}}\\ \end{split} \end{equation}

Plug in 3 and 4 in each of 2 fractions:

\begin{equation} \begin{split} \\&= \frac{\color{red}{3^2}}{1+1^\color{red}{3}}+ \frac{\color{red}{3^2}}{1+2^\color{red}{3}}+ \\&+\frac{\color{red}{4^2}}{1+1^\color{red}{4}}+ \frac{\color{red}{4^2}}{1+2^\color{red}{4}} \\&=\frac{9}{2}+\frac{9}{9}+\frac{16}{2}+\frac{16}{17} \\&=\frac{1377+306+2448+288}{306} \\&=\frac{4419}{306} \\&=\frac{491}{34} \end{split} \end{equation}

Now it is your turn to practice to become more confident in double summation. Many examples to practice are coming!

If any problem, look at the STEP 1 and STEP 2, and at the solved examples.
You can check the solution at the bottom of the page as usual.

Double summation with one summation variable

Now we will look at the double summation with one summation variable. So we will have 2 summation indexes (i and j, for example), but next to two summation signs, there will be only one variable (i or j).

Example:

\begin{equation} \begin{split} \sum_{{i=1}}^{5}\sum_{{j=1}}^{4} j& \end{split} \end{equation}

Solution: We solve this example in the same way as the previous ones.

So first, we will sum the inner sum with summation variable j:

\begin{equation} \begin{split} \sum_{{i=1}}^{5}\sum_{{j=1}}^{4} j&=\sum_{{i=1}}^{5} (1+2+3+4) \\&=\sum_{{i=1}}^{5} 10 \end{split} \end{equation}

\begin{equation} \begin{split} &\sum_{{i=1}}^{5}\sum_{{j=1}}^{4} j=\\&=\sum_{{i=1}}^{5} (1+2+3+4) \\&=\sum_{{i=1}}^{5} 10 \end{split} \end{equation}

Now we have a summation of a constant. Since there is no summation variable 'i' next to the summation sign (so everything else except 'i' is treated as a constant).

And second, we add the constant 10 five times, as we learned from the previous sections:

\begin{equation} \begin{split} \sum_{{i=1}}^{5}\sum_{{j=1}}^{4} j&=\sum_{{i=1}}^{5} (1+2+3+4) \\&=\sum_{{i=1}}^{5} 10 \\&=10 +10+10+10+10 \\&=50 \end{split} \end{equation}

\begin{equation} \begin{split} &\sum_{{i=1}}^{5}\sum_{{j=1}}^{4} j=\\&=\sum_{{i=1}}^{5} (1+2+3+4) \\&=\sum_{{i=1}}^{5} 10 \\&=10 +10+10+10+10 \\&=50 \end{split} \end{equation}

The end of the example

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Let's have another one but this time with variable "i". Then you will solve a couple yourself.

Example:

\begin{equation} \begin{split} \sum_{{i=1}}^{5}\sum_{{j=1}}^{4} i^2& \end{split} \end{equation}

Solution: Let's solve the inner sum first. The summation index is 'j,' but we don't have it next to a summation sign. Therefore, we will add $i^2$ four times, as it is treated as a constant, and the index 'j' goes from 1 to 4 (see Summation of a constant)
So:

\begin{equation} \begin{split} \sum_{{i=1}}^{5}\sum_{{j=1}}^{4} i^2&=\sum_{{i=1}}^{5} 4\cdot i^2\\ \\ and\hspace{0.4em} now \hspace{0.4em} we\\\hspace{0.4em}solve\hspace{0.4em} as\hspace{0.4em} usual: \\&= 4\cdot 1^2 +4\cdot 2^2+ \\&+4\cdot 3^2+4\cdot 4^2+ 4\cdot 5^2 \\&=4\cdot (1+4+9+16+25) \\&=4\cdot 55 \\&=220 \end{split} \end{equation}

\begin{equation} \begin{split} &\sum_{{i=1}}^{5}\sum_{{j=1}}^{4} i^2=\\&=\sum_{{i=1}}^{5} 4\cdot i^2 \\&=\sum_{{i=1}}^{5} 4\cdot 1^2 +4\cdot 2^2+ \\&+4\cdot 3^2+4\cdot 4^2+ 4\cdot 5^2 \\&=4\cdot (1+4+9+16+25) \\&=4\cdot 55 \\&=220 \end{split} \end{equation}

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And the last one together. Then it will be your turn!

Let's solve together:

\begin{equation} \begin{split} \sum_{{i=5}}^{7}\sum_{{j=10}}^{15} i^{2\cdot i} \end{split} \end{equation}

So, during the first step, we will solve the inner sum as usual. Again, $i^{2\cdot i}$ is just a constant since there isn't any 'j' variable. So, how many times do we add $i^{2\cdot i}$? We add it as many times as there are numbers between 10 and 15, including both 10 and 15. Therefore, it's $15-10+1=6$ times (check: 10, 11, 12, 13, 14, 15).
So we have:

\begin{equation} \begin{split} \sum_{{i=5}}^{7}\sum_{{j=10}}^{15} i^{2\cdot i}&= \sum_{{i=5}}^{7} 6\cdot i^{2\cdot i} \\and\hspace{0.4em}now\hspace{0.4em}as\hspace{0.4em}usual \\&=6\cdot 5^{2\cdot 5}+6\cdot 6^{2\cdot 6}+6\cdot 7^{2\cdot 7} \\&=6\cdot( 5^{10}+6^{12}+7^{14}) \\thats\hspace{0.4em}huge \\&=4082457724860 \end{split} \end{equation}

\begin{equation} \begin{split} &\sum_{{i=5}}^{7}\sum_{{j=10}}^{15} i^{2\cdot i}=\\&= \sum_{{i=5}}^{7} 6\cdot i^{2\cdot i} \\&=6\cdot 5^{2\cdot 5}+6\cdot 6^{2\cdot 6}+6\cdot 7^{2\cdot 7} \\&=6\cdot( 5^{10}+6^{12}+7^{14}) \\&=4082457724860 \end{split} \end{equation}

Now it's your time to shine. Answers as always are at the bottom of the page. You have everything necessery to solve them. If any problem look on examples we solved together or follow the links to previous topics mentioned on this page.

What about if there isn't any index variable next to a summation sign?

Double summation of a constant (no variable)

Let's consider following double summation example with index variables "i" and "j":

Example:

\begin{equation} \begin{split} \sum_{{i=1}}^{5}\sum_{{j=1}}^{4} 5k& \end{split} \end{equation}

Solution: as you can see, we have only constant next to summation signs - 5k.
Meaning that we add 5k four times (as j goes from 1 to 4), getting:

1. $\sum_{{i=1}}^{5}\sum_{{j=1}}^{4} 5k=\sum_{{i=1}}^{5}4\cdot 5k=\sum_{{i=1}}^{5}20k$

1. $\sum_{{i=1}}^{5}\sum_{{j=1}}^{4} 5k=\sum_{{i=1}}^{5}4\cdot 5k= \\ =\sum_{{i=1}}^{5}20k$

2. Now we solve single summation of a constant as usual:
$\sum_{{i=1}}^{5}20 k=5\cdot 20k=100k$

Final answer: $\sum_{{i=1}}^{5}\sum_{{j=1}}^{4} 5k=100k$

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Constants can be anything, not just letters. When there is a constant in a double summation, we simply need to add it the correct number of times:

Let's solve together:

\begin{equation} \begin{split} \sum_{{i=1}}^{5}\sum_{{j=1}}^{6} 😃& \end{split} \end{equation}

Solution: first we add emoji 6x:

$\sum_{{i=1}}^{5}\sum_{{j=1}}^{6} 😃=\sum_{{i=1}}^{5}6\cdot 😃$

And then 5x again:

$\sum_{{i=1}}^{5}6\cdot 😃=5\cdot 6\cdot 😃=30\cdot 😃$

Remember! Everything that is not the index variable is a constant for the given summation sign. Is it important to start from the inner sum? No, you can start from the outer one as well, as it is not that hard to see. Check it yourself.

Solutions

The sums are: