Given the ubiquity of summations in mathematics,
let's find a way to abbreviate this operation
by using sigma notation.
In this section we discuss the following:
For any constant \(c,\) show that
\[\sum_{i = 1}^n c = c \, n \pd\]
We see
\[
\ba
\sum_{i = 1}^n c &= \underbrace{c + c + c + \cdots}_{n \text{ terms}} \nl
&= c \, n \cma
\ea
\]
as requested.
EXAMPLE 3
Prove that
\[\sum_{i = 1}^n i = \frac{n(n + 1)}{2} \pd\]
Let's write the terms in ascending order and then in descending order, as follows:
\[
\baat{7}
S &= 1 &&+ 2 &&+ 3 &&+ \cdots &&+ (n - 2) &&+ (n - 1) &&+ n \cma \nl
S &= n &&+ (n - 1) &&+ (n - 2) &&+ \cdots &&+ 3 &&+ 2 &&+ 1 \pd
\eaat
\]
Adding every column vertically gives
\[
\ba
2S &= \underbrace{(n + 1) + (n + 1) + (n + 1) + (n + 1) + \cdots}_{n \textrm{ terms}} \nl
&= n(n + 1) \pd
\ea
\]
Then solving for \(S\) gives, as requested,
\[S = \frac{n(n + 1)}{2} \pd\]
EXAMPLE 4
Using a Proof by Induction, prove that
\[\sum_{i = 1}^n i^2 = \frac{n(n + 1)(2n + 1)}{6} \pd\]
Let \(S_n\) be the statement
\[\sum_{i = 1}^n i^2 = \frac{n(n + 1)(2n + 1)}{6} \pd\]
To perform a Proof by Induction,
we first show that \(S_1\) is true and then show that
if \(S_k\) is true (where \(k\) is a positive integer), then \(S_{k + 1}\) is also true.
First observe that
\[S_1 = 1^2 \equalsCheck \frac{1(1 + 1)(2 + 1)}{6} \pd\]
Next assume that \(S_k\) is true—that is,
\[\sum_{i = 1}^k i^2 = \frac{k(k + 1)(2k + 1)}{6} \pd\]
Observe that
\[
\ba
S_{k + 1} &= \sum_{i = 1}^{k + 1} i^2 \nl
&= \sum_{i = 1}^k i^2 + (k + 1)^2 \nl
&= \frac{k(k + 1)(2k + 1)}{6} + (k + 1)^2 \nl
&= (k + 1) \parbr{\frac{k (2k + 1)}{6} + (k + 1)} \nl
&= (k + 1) \parbr{\frac{k (2k + 1) + 6(k + 1)}{6}} \nl
&= (k + 1) \parbr{\frac{2k^2 + 7k + 6}{6}} \nl
&= (k + 1) \parbr{\frac{(k + 2)(2k + 3)}{6}} \nl
&= \frac{(k + 1) [(k + 1) + 1] [2(k + 1) + 1]}{6} \pd
\ea
\]
Thus, \(S_{k + 1}\) is also true.
By mathematical induction, \(S_n\) is true
for all positive integers \(n.\)
Properties of Summations
Instead of performing long computations, let's use rules to
simplify the evaluation of summations.
PROPERTIES OF SUMMATIONS
If \(c\) is a constant, and \(a_i\) and \(b_i\) are real numbers for all \(m \leq i \leq n,\)
then we have the following properties:
\begin{align}
\sum_{i = m}^n (a_i + b_i) &= \sum_{i = m}^n a_i + \sum_{i = m}^n b_i \cma \label{eq:sigma-sum} \nl
\sum_{i = m}^n (a_i - b_i) &= \sum_{i = m}^n a_i - \sum_{i = m}^n b_i \cma \label{eq:sigma-diff} \nl
\sum_{i = m}^n c a_i &= c \sum_{i = m}^n a_i \pd \label{eq:sigma-cons} \nl
\end{align}
In words, the summation of a sum is the sum of summations [by \(\eqref{eq:sigma-sum}\)] and
the summation of a difference is the difference of summations [by \(\eqref{eq:sigma-diff}\)].
By \(\eqref{eq:sigma-cons},\) we can pull out a constant from the summation expression.
Let's also provide a list of useful formulas,
which include the results of Example 2,
Example 3,
and
Example 4.
SUMMATION FORMULAS
Let \(c\) be a constant.
\begin{align}
\sum_{i = 1}^n c &= c \, n \cma \label{eq:sum-c} \nl
\sum_{i = 1}^n i &= \frac{n(n + 1)}{2} \cma \label{eq:sum-i} \nl
\sum_{i = 1}^n i^2 &= \frac{n(n + 1)(2n + 1)}{6} \cma \label{eq:sum-i2} \nl
\sum_{i = 1}^n i^3 &= \parbr{\frac{n(n + 1)}{2}}^2 \pd \label{eq:sum-i3}
\end{align}
EXAMPLE 5
\[\sum_{i = 1}^{10} \par{2i^3 - 6i^2 + 8i}\]
Let the sum be \(S.\)
We decompose the series one step at a time, as follows:
\[
\baat{2}
S
&= \sum_{i = 1}^{10} 2i^3 - \sum_{i = 1}^{10} 6 i^2 + \sum_{i = 1}^{10} 8i
&&\comment{\text{by } \eqref{eq:sigma-sum} \text{ and } \eqref{eq:sigma-diff}} \nl
&= 2 \sum_{i = 1}^{10} i^3 - 6 \sum_{i = 1}^{10} i^2 + 8 \sum_{i = 1}^{10} i
&&\comment{\text{by } \eqref{eq:sigma-cons}} \nl
&= 2 \parbr{\frac{10(10 + 1)}{2}}^2 - 6 \parbr{\frac{10(10 + 1)(2 \cdot 10 + 1)}{6}} + 8 \parbr{\frac{10(10 + 1)}{2}}
&&\comment{\text{by } \eqref{eq:sum-i3} \cma \eqref{eq:sum-i2} \cma \text{and } \eqref{eq:sum-i}} \nl
&= \boxed{4180}
\eaat
\]
Defining Sigma Notation
If \(a_m,\) \(a_{m + 1},\) \(a_{m + 2}, \dots,\) \(a_{n - 1},\) \(a_n\)
are real numbers, and \(m\) and \(n\) are integers such that \(m \leq n,\)
then
\begin{equation}
\sum_{i = m}^n a_i = a_m + a_{m + 1} + a_{m + 2} + \cdots + a_{n - 1} + a_n \pd \eqlabel{eq:sigma-def}
\end{equation}
Properties of Summations
If \(c\) is a constant, and \(a_i\) and \(b_i\) are real numbers for all \(m \leq i \leq n,\)
then we have the following properties:
\begin{align}
\sum_{i = m}^n (a_i + b_i) &= \sum_{i = m}^n a_i + \sum_{i = m}^n b_i \cma \eqlabel{eq:sigma-sum} \nl
\sum_{i = m}^n (a_i - b_i) &= \sum_{i = m}^n a_i - \sum_{i = m}^n b_i \cma \eqlabel{eq:sigma-diff} \nl
\sum_{i = m}^n c a_i &= c \sum_{i = m}^n a_i \pd \eqlabel{eq:sigma-cons} \nl
\end{align}
The following formulas also prove useful in dissecting sums:
\begin{align}
\sum_{i = 1}^n c &= c \, n \cma \eqlabel{eq:sum-c} \nl
\sum_{i = 1}^n i &= \frac{n(n + 1)}{2} \cma \eqlabel{eq:sum-i} \nl
\sum_{i = 1}^n i^2 &= \frac{n(n + 1)(2n + 1)}{6} \cma \eqlabel{eq:sum-i2} \nl
\sum_{i = 1}^n i^3 &= \parbr{\frac{n(n + 1)}{2}}^2 \pd \eqlabel{eq:sum-i3}
\end{align}
