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Section 7-8 : Summation Notation

In this section we need to do a brief review of summation notation or sigma notation. We’ll start out with two integers, \(n\) and \(m\), with \(n < m\) and a list of numbers denoted as follows,

\[{a_n},\,\,{a_{n + 1}},\,\,{a_{n + 2}},\,\, \ldots ,\,\,{a_{m - 2}},\,\,{a_{m - 1}},\,\,{a_m}\]

We want to add them up, in other words we want,

\[{a_n} + \,{a_{n + 1}} + \,\,{a_{n + 2}} + \,\, \ldots + \,\,{a_{m - 2}} + \,\,{a_{m - 1}} + \,\,{a_m}\]

For large lists this can be a fairly cumbersome notation so we introduce summation notation to denote these kinds of sums. The case above is denoted as follows.

\[\sum\limits_{i = n}^m {{a_i}} = {a_n} + \,{a_{n + 1}} + \,\,{a_{n + 2}} + \,\, \ldots + \,\,{a_{m - 2}} + \,\,{a_{m - 1}} + \,\,{a_m}\]

The \(i\) is called the index of summation. This notation tells us to add all the \({a_i}\)’s up for all integers starting at \(n\) and ending at \(m\).

For instance,

\[\begin{align*}\sum\limits_{i = 0}^4 {\frac{i}{{i + 1}}} & = \frac{0}{{0 + 1}} + \frac{1}{{1 + 1}} + \frac{2}{{2 + 1}} + \frac{3}{{3 + 1}} + \frac{4}{{4 + 1}} = \frac{{163}}{{60}} = 2.7166\overline{6}\\ \sum\limits_{i = 4}^6 {{2^i}{x^{2i + 1}}} & = {2^4}{x^9} + {2^5}{x^{11}} + {2^6}{x^{13}} = 16{x^9} + 32{x^{11}} + 64{x^{13}}\\ \sum\limits_{i = 1}^4 {f\left( {x_i^*} \right)} & = f\left( {x_1^*} \right) + f\left( {x_2^*} \right) + f\left( {x_3^*} \right) + f\left( {x_4^*} \right)\end{align*}\]

Properties

Here are a couple of formulas for summation notation.

  1. \(\sum\limits_{i\, = \,{i_{\,0}}}^n {c{a_i}} = c\sum\limits_{i\, = \,{i_{\,0}}}^n {{a_i}} \) where \(c\) is any number. So, we can factor constants out of a summation.

  2. \(\sum\limits_{i\, = \,{i_{\,0}}}^n {\left( {{a_i} \pm {b_i}} \right)} = \sum\limits_{i\, = \,{i_{\,0}}}^n {{a_i}} \pm \sum\limits_{i\, = \,{i_{\,0}}}^n {{b_i}} \) So, we can break up a summation across a sum or difference.

Note that we started the series at \({i_{\,0}}\) to denote the fact that they can start at any value of \(i\) that we need them to. Also note that while we can break up sums and differences as we did in 2 above we can’t do the same thing for products and quotients. In other words,

\[\sum\limits_{i\, = \,{i_{\,0}}}^n {\left( {{a_i}{b_i}} \right)} \ne \left( {\sum\limits_{i\, = \,{i_{\,0}}}^n {{a_i}} } \right)\left( {\sum\limits_{i\, = \,{i_{\,0}}}^n {{b_i}} } \right)\hspace{0.75in}\sum\limits_{i\, = \,{i_{\,0}}}^n {\frac{{{a_i}}}{{{b_i}}}} \ne \frac{{\sum\limits_{i\, = \,{i_{\,0}}}^n {{a_i}} }}{{\sum\limits_{i\, = \,{i_{\,0}}}^n {{b_i}} }}\]

Formulas

Here are a couple of nice formulas that we will find useful in a couple of sections. Note that these formulas are only true if starting at \(i = 1\). You can, of course, derive other formulas from these for different starting points if you need to.

  1. \(\displaystyle \sum\limits_{i = 1}^n c = cn\)

  2. \(\displaystyle \sum\limits_{i = 1}^n i = \frac{{n\left( {n + 1} \right)}}{2}\)

  3. \(\displaystyle \sum\limits_{i = 1}^n {{i^2}} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\)

  4. \(\displaystyle \sum\limits_{i = 1}^n {{i^3}} = {\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]^2}\)

Here is a quick example on how to use these properties to quickly evaluate a sum that would not be easy to do by hand.

Example 1 Using the formulas and properties from above determine the value of the following summation. \[\sum\limits_{i = 1}^{100} {{{\left( {3 - 2i} \right)}^2}} \]
Show Solution

The first thing that we need to do is square out the stuff being summed and then break up the summation using the properties as follows,

\[\begin{align*}\sum\limits_{i = 1}^{100} {{{\left( {3 - 2i} \right)}^2}} & = \sum\limits_{i = 1}^{100} {9 - 12i + 4{i^2}} \\ & = \sum\limits_{i = 1}^{100} 9 - \sum\limits_{i = 1}^{100} {12i} + \sum\limits_{i = 1}^{100} {4{i^2}} \\ & = \sum\limits_{i = 1}^{100} 9 - 12\sum\limits_{i = 1}^{100} i + 4\sum\limits_{i = 1}^{100} {{i^2}} \end{align*}\]

Now, using the formulas, this is easy to compute,

\[\begin{align*}\sum\limits_{i = 1}^{100} {{{\left( {3 - 2i} \right)}^2}} & = 9\left( {100} \right) - 12\left( {\frac{{100\left( {101} \right)}}{2}} \right) + 4\left( {\frac{{100\left( {101} \right)\left( {201} \right)}}{6}} \right)\\ & = 1293700\end{align*}\]

Doing this by hand would definitely taken some time and there’s a good chance that we might have made a minor mistake somewhere along the line.